Factorization of the hypergeometric-type difference equation on the non-uniform lattices: dynamical algebra

Factorization of the hypergeometric-type difference equation on the non-uniform lattices: dynamical algebra R. ´ Alvarez-Nodarse†‡, N. M. Atakishiyev§and R. S. Costas-Santos∗ †Departamento de An´alisis Matem´atico. Universidad de Sevilla. Apdo. 1160, E-41080 Sevilla, Spain ‡Instituto Carlos I de F´ısica Te´orica y Computacional, Universidad de Granada, E-18071 Granada, Spain §Instituto de Matem´aticas, UNAM, Apartado Postal 273-3, C.P. 62210 Cuernavaca, Morelos, M´exico ∗Departamento de Matem´aticas, E.P.S., Universidad Carlos III de Madrid. Ave. Universidad 30, E-28911, Legan´es, Madrid, Spain 14th October 2004 Abstract We argue that one can factorize the difference equation of hypergeometric type on the nonuniform lattices in general case. It is shown that in the most cases of q-linear spectrum of the eigenvalues this directly leads to the dynamical symmetry algebra suq(1,1), whose generators are explicitly constructed in terms of the difference operators, obtained in the process of factorization. Thus all models with the q-linear spectrum (some of them, but not all, previously considered in a number of publications) can be treated in a unified form. 1 Introduction and preliminaries In this paper we continue the study, started in [1], on the factorization of the hypergeometrictype difference equation on the non-uniform lattices, i.e., of the equation [2] σ(s)∆ ∆x(s−1 2)∇y(s) ∇x(s)+τ(s)∆y(s) ∆x(s)+λy(s) = 0, σ(s) = ˜σ(x(s)) −1 2˜τ(x(s))∆xs−1 2, τ(s) = ˜τ(x(s)), (1) where ∆y(s) := y(s+ 1) −y(s), ∇y(s) := y(s)−y(s−1), ˜σ(x(s)) and ˜τ(x(s)) are polynomials in x(s) of degree at most 2 and 1, respectively, and λis a constant (see also [3]). The difference equation (1) has polynomial solutions Pn(x(s); q):= Pn(s;q) of the hypergeometric type if and only if the lattice x(s) has the form [4, 5] x(s) = c1(q)qs+c2(q)q−s+c3(q) = c1(q)[qs+q−s−µ] + c3(q),(2) where c1,c2,c3and qµ:= c1/c2are constants which, in general, depend on q. An important special case of the lattice x(s) is the q-linear lattice, which is obtained from (2) by assuming that either c1(q) or c2(q) vanishes. 1 The polynomial solutions of the difference equation (1) correspond to the following expression [3] for its eigenvalues λn(q): λn(q) = C1qn+C2q−n+C3, C1=1 2(1 −q)eτ0+eσ00 kq, C2=1 2(1 −q−1)eτ0−eσ00 kq, C3=−eσ00(1 + q) 2kq(1 −q)−eτ0 2, (3) where eτ0and eσ00 are the coefficients of x(s) and x2(s) in the Taylor expansion for ˜τ(x(s)) and ˜σ(x(s)), respectively, i.e., ˜τ(x(s)) = eτ0x(s) + eτ(0), and ˜σ(x(s)) = eσ00/2x2(s) + eσ0(0)x(s) + eσ(0). Observe that the coefficients C1and C2of the qnand q−nterms, respectively, are fixed by the functions σand τin (1), and so is the product C1C2. In what follows we denote by Lqthe value of C1C2=(eσ00/kq)2−(eτ0)2/4k2 q. The sequence {λn(q)}satisfies the following three-term recurrence relation (TTRR) λn+2(q)−(q+q−1)λn+1(q) + λn(q) = 1 2(˜τ0k2 q−˜σ00[2]q) = C. (4) Conversely, if {λn(q)}satisfy the TTRR (4), then it has the form λn(q) = C0 1qn+C0 2q−n+C0 3. Obviously, having used the initial conditions λ0(q) = 0 and λ1(q) = −eτ0, one recovers the expression (3). It is well known [3] that under certain conditions the polynomial solutions of (1) are orthogonal. For example, if σ(s)ρ(s)xk(s−1 2)s=a,b = 0, for all k= 0,1,2,..., then the polynomial solutions possess a discrete orthogonality property b−1 X s=a Pn(s;q)Pm(s;q)ρ(s)∇x1(s) = d2 n(q)δn,m,(5) where the weight function ρ(s) is a solution of the Pearson-type difference equation [3] ∆ ∆x(s−1 2)[σ(s)ρ(s)] = τ(s)ρ(s) or σ(s+ 1)ρ(s+ 1) = σ(−s−µ)ρ(s).(6) If the lattice x(s) is a q-linear lattice, i.e., x(s) = c q±s+c3, then the σ(−s−µ) in (6) should be substituted by σ(s) + τ(s)∆x(s−1/2). A more detailed information on orthogonal polynomials on the non-uniform lattices can be found in [3, 5, 6, 7, 8]. In [1] it has been shown that one can factorize the Nikiforov-Uvarov equation (1) with the aid of raising and lowering operators, which can be constructed for solutions of this equation. In this paper we wish to make one step further by studying the dynamical symmetry algebra for the hypergeometric-type difference equation (1) on the non-uniform lattices (2). Our approach is essentially based on the simple observation, formulated in [9]: In order to factorize an arbitrary difference equation, one should express it explicitly in terms of the shift (or displacement) operators exp(ad ds), which are defined as exp(ad ds)f(s) = f(s+a), ais some constant. For example, in the case of the equation (1) this corresponds to the substitutions ∆ = exp( d ds )−1 and ∇= 1 −exp(−d ds). This procedure converts a difference equation into an eigenvalue problem for a difference operator, represented by a linear combination of some shift operators (with coefficients, which depend polynomially on the variable s). Since each term of this linear combination is readily factorizable (because exp (α+β)A= exp α A exp β A for an arbitrary operator A), the factorization of the whole linear combination, which represents the 2 initial difference equation, becomes straightforward. Inspired by the appearance of Macfarlane’s [10] and Biedenharn’s [11] important constructions of q-analogues of quantum harmonic oscillator, this technique of factorization of difference equations was later employed in a number of publications [12]–[16] in order to study group theoretic properties of the various well-known families of orthogonal polynomials, which can be viewed as q-extensions of the classical Hermite polynomials. So our purpose here is to formulate a unified approach to deriving all of these results, which correspond to the q-linear spectrum. An important aspect to observe at this point is that we shall mainly (except for the examples in subsection 4.2) confine our attention to those families of q-polynomials, which satisfy discrete orthogonality relation of the type (5). The explanation of such preference is that the factorization of difference equations for instances of q-polynomials with continuous orthogonality property has been already thoroughly studied in [12]–[16]. Observe also that our approach still remains valid in the limit as q→1; so classical counterparts of q-polynomials, which will be discussed in this paper, are in fact incorporated as appropriate limit cases. But the reader who desires to know more about the factorization in the cases of classical orthogonal polynomials (such as the Kravchuk, Charlier, Meixner, Meixner–Pollaczek, and Hahn) may be referred to [17, 18] and references therein. The paper is organized as follows. In section 2 we associate with each family of q-polynomials a “q-Hamiltonian” H(s;q) (via the second-order difference equation) and construct two difference operators a(s;q) and b(s;q), which factorize the operator H(s;q). Our main results are given in section 3: they are formulated in Theorems 3.4 – which gives a simple necessary and sufficient condition that the q-Hamiltonian H(s;q) admits the factorization in terms of the operators a(s;q) and b(s;q), which satisfy the relation a(s;q)b(s;q)−qγb(s;q)a(s;q) = Ifor some γ, and 3.5 – stating that the eigenvalues of the difference equation (1) in this case should be of the form λn(q) = C1qn+C3or λn(q) = C2q−n+C3. In section 4 several relevant examples of particular q-families of orthogonal polynomials are illustrated. 2 Factorization operators Let introduce a set of functions Φn Φn(s;q) = d−1 nA(s)pρ(s)Pn(s;q),(7) where dnis the norm of the q-polynomials Pn(s;q), ρ(s) is the solution of the Pearson equation (6) and A(s) is an arbitrary continuous function, A(s)6= 0 in the interval (a, b) of orthogonality of Pn. If the polynomials Pn(s;q) possess the discrete orthogonality property (5), then the functions Φn(s;q) satisfy hΦn(s;q),Φm(s;q)i= b−1 X s=a Φn(s;q)Φm(s;q)∇x1(s) A2(s)=δn,m.(8) Notice that if A(s) = p∇x1(s), then the set (Φn)nis an orthonormal set. Obviously, in the case of a continuous orthogonality (as for the Askey-Wilson polynomials) one needs to change the sum in (8) by a Riemann integral [3, 5]. Next, we define the q-Hamiltonian H(s;q) of the form H(s;q) := 1 ∇x1(s)A(s)H(s;q)1 A(s),(9) 3 where H(s;q) := −pσ(−s−µ+1)σ(s) ∇x(s)e−∂s−pσ(−s−µ)σ(s+ 1) ∆x(s)e∂s+σ(−s−µ) ∆x(s)+σ(s) ∇x(s)I, (10) eα∂sf(s) = f(s+α) for all α∈Cand Iis the identity operator. If we now use the identity ∇= ∆ −∇∆ and the equation (1), we find that H(s;q)Φn(s;q) = λnΦn(s;q),(11) i.e., the functions Φn(s;q), defined in (7), are the eigenfunctions of the associated operator H(s;q). Our first step is to find two operators a(s;q) and b(s;q) such that the Hamiltonian H(s;q) = b(s;q)a(s;q), i.e., the operators a(s;q) and b(s;q)factorize the q-Hamiltonian H(s;q). But before exhibiting their explicit form let us point out that if there exists a pair of such operators, then there are infinitely many of them. Indeed, let a(s;q) and b(s;q) be such operators that H(s;q) = b(s;q)a(s;q) and let U(s;q) be an arbitrary unitary operator, i.e., U†(s;q)U(s;q) = I. Then the operators ea(s;q) := U(s;q)a(s;q),eb(s;q) := b(s;q)U†(s;q), also factorize H(s;q) for eb(s;q)ea(s;q) = b(s;q)U†(s;q)U(s;q)a(s;q) = b(s;q)I a(s;q) = b(s;q)a(s;q) = H(s;q). This arbitrariness in picking up a particular unitary operator U(s) is very essential because it enables one to construct a closed algebra, which contains a Hamiltonian H(s;q) itself. An explicit form of the spectrum of this Hamiltonian may then be found by purely algebraic arguments from the knowledge of representations of this algebra (which is therefore referred to as a dynamical algebra). If one applies the standard factorization procedure to the equation (1), then the following difference operators emerge Definition 2.1 Let αbe a real number and A(s)an arbitrary continuous non-vanishing function in (a, b). We define a family of α-down and α-up operators by a↓ α(s;q):= A(s) p∇x1(s)e−α∂s e∂ssσ(s) ∇x(s)−sσ(−s−µ) ∆x(s)!1 A(s), a↑ α(s;q):= 1 ∇x1(s)A(s) sσ(s) ∇x(s)e−∂s−sσ(−s−µ) ∆x(s)!eα∂sp∇x1(s) A(s), (12) respectively. A straightforward calculation (by using the simple identity e∂s∇= ∆) shows that for all α∈R H(s;q) = a↑ α(s;q)a↓ α(s;q),(13) i.e., the operators a↓ α(s;q) and a↑ α(s;q) factorize the Hamiltonian, defined in (9). i.e., we have the following Theorem 2.2 Given a q-Hamiltonian (9) H(s;q), then the operators a↓ α(s;q)and a↑ α(s;q)defined in (12) are such that for all α∈C,H(s;q) = a↑ α(s;q)a↓ α(s;q). 4 Our next step is to find a dynamical algebra, associated with the Hamiltonian H(s;q). To this end we will need the following definition Definition 2.3 A function f(z)is said to be a linear-type function of z, if there exist two functions Fand G, such that for all z, ζ ∈C, this function f(z)can be represented as f(z+ζ) = F(ζ)f(z) + G(ζ). A particular case of the linear-type functions are the q-linear functions, i.e., the functions of the form f(z) = Aqz+B. For these functions f(z+ζ) = F(ζ)f(z) + G(ζ), where F(ζ) = qζand G(ζ) = B(1 −qζ). Remark 2.4 If we use the expression in (3) for the eigenvalues λn, then it is straightforward to see that λnis a q-linear function of nif and only if eσ00 =±kqeτ0. Moreover, in this case we have eσ00 =kqeτ0⇒λn(q, +) = eτ0 1−q(qn−1),and eσ00 =−kqeτ0⇒λn(q, −) = eτ0 1−q−1(q−n−1). (14) Notice that λn(q, −) = λn(q−1,+), i.e., the second case can be obtained form the first one just by changing qto q−1. Proposition 2.5 The function λnis a q-linear function of nif and only if it satisfies λn+1 = qλn+C. Proof: A straightforward computations show that if λnis a q-linear function of n, then it satisfies the recurrence formula λn+1 =qλn+C, where Cis a constant (in this case C=λ1). But the general solution of the difference equation λn+1 =qλn+Cis λn=Aqn+D, where A and Dare, in general, non-vanishing constants. Remark 2.6 Notice that if λnis a q-linear function of n, then λnsatisfies the recurrence relation λn+γ−qγλn=Cfor any numbers γand C. Finally, we have the following straightforward lemma Lemma 2.7 Let x(s)be a q-linear function of sand λnbe the eigenvalue of the difference equation of hypergeometric type (1). Then λnis a q-linear function of nif and only if ∆(2)(σ(s)) = 0 and q−1-linear function of nif and only if ∆(2)(σ(−s−µ)) = 0, where ∆(2) is the operator ∆(2) =∆ ∆x1(s) ∆ ∆x(s). Proof: It follows from equation (14) and the fact that ∆(2)(σ(s)) = [2]q 2(eσ00 −˜τ0kq) and ∆(2)(σ(−s−µ)) = [2]q 2(eσ00 + ˜τ0kq). 3 Dynamical algebra We begin this section with the following definition. Definition 3.1 Let ςbe a real number, and let a(s;q)and b(s;q)be two operators. We define the ς-commutator of aand bas [a(s;q), b(s;q)]ς=a(s;q)b(s;q)−ςb(s;q)a(s;q). 5 Proposition 3.2 Let H(s;q)be an operator, such that there exist two operators a(s;q)and b(s;q)and two real numbers ςand Λ, such that H(s;q) = b(s)a(s;q), and [a(s;q), b(s;q)]ς= Λ. Then, if Φ(s;q)is an eigenvector of the Hamiltonian H(s;q), associated with the eigenvalue λ, we have 1. H(s;q){a(s;q)Φ(s;q)}=ς−1(λ−Λ) {a(s;q)Φ(s;q)}, i.e., a(s;q)Φ(s;q)is the eigenvector of H(s;q), associated with the eigenvalue ς−1(λ−Λ), 2. H(s;q){b(s;q)Φ(s;q)}= (Λ + ςλ){b(s;q)Φ(s;q)}, i.e., b(s;q)Φ(s;q)is the eigenvector of H(s;q), associated with the eigenvalue Λ + ςλ. Proof: In the first case, since H(s;q)Φ(s;q) = λΦ(s;q), H(s;q){a(s;q)Φ(s;q)}=b(s;q)a(s;q){a(s;q)Φ(s;q)}=ς−1(a(s;q)b(s;q)−Λ){a(s;q)Φ(s;q)} =ς−1(λ−Λ){a(s;q)Φ(s;q)}. By the same token, in the second case H(s;q){b(s;q)Φ(s;q)}=b(s;q)a(s;q)b(s;q)Φ(s;q) = b(s;q)(Λ + ςλ)Φ(s;q) = (Λ + ςλ){b(s;q)Φ(s;q)}. In the same way one can prove that a(s;q)b(s;q)Φ(s;q) = (Λ + ςλ)Φ(s;q).(15) Moreover, if Φ(s;q) is an eigenvector of the Hamiltonian H(s;q) (or of the operator a(s;q)b(s;q)), then ak(s;q)Φ(s;q) and bk(s;q)Φ(s;q) are, in general, also eigenvectors. Remark 3.3 Obviously, the condition [a(s;q), b(s;q)]ς=Ican be changed to [a(s;q), b(s;q)]ς= Λ, where Λis an arbitrary non-zero constant. In fact, if the operators a(s;q)and b(s;q)satisfy the q-commutation relation [a(s;q), b(s;q)]ς= Λ, then the operators a(s;q) = Λ−1/2a(s;q)and b(s;q) = Λ−1/2b(s;q)satisfy [a(s;q),b(s;q)]ς=I, and H(s;q) = Λb(s;q)a(s;q). The Proposition 3.2 thus refers to the case of a system, described by a Hamiltonian H(s;q), which admits the factorization (13) in terms of the operators a(s;q) and b(s;q), satisfying the q-commutation relation [a(s;q), b(s;q)]ς=I. Moreover, it tells us how to construct a dynamical symmetry algebra for such a case in a direct fashion [19]. Indeed, let us assume that [a(s;q), b(s;q)]ς=I, ς =q2(or q−2), and b(s;q) = a†(s;q). Then one can rewrite the q2commutator a(s;q)a†(s;q)−q2a†(s;q)a(s;q) = Iin the following form [a(s;q), a†(s;q)] := a(s;q)a†(s;q)−a†(s;q)a(s;q) = I−(1 −q2)a†(s;q)a(s;q) := q2N(s), where, by definition, the operator N(s) is equal to N(s) = ln[I−(1 −q2)a†(s;q)a(s;q)]/ln q2. From this definition of N(s) it follows that [N(s), a(s;q)] = −a(s;q),[N(s), a†(s;q)] = a†(s;q),(16) i.e., N(s) is the number operator. The next (and final) step is to introduce a new set of the operators b(s;q) := q−N(s)/2a(s;q), b†(s;q) := a†(s;q)q−N(s)/2, which satisfy the following commutation relation b(s;q)b†(s;q)−q b†(s;q)b(s;q) = q−N(s), 6 readily derived with the aid of (16). The operators b(s;q), b†(s;q), and N(s) directly lead to the dynamical algebra suq(1,1) with the generators K0(s) = 1 2[N(s) + 1/2], K+(s) = β(b†(s;q))2, K−(s) = β b2(s;q), β−1=q+q−1. It is straightforward to verify that thus defined generators satisfy the standard commutation relations [K0(s), K±(s)] = ±K±(s),[K−(s), K+(s)] = [2K0(s)]q2, of the algebra suq(1,1) (see e.g. [20]). Thus in the case when the operators a(s;q) and b(s;q), which factorize the Hamiltonian H(s;q), satisfy the q-commutation relation [a(s;q), b(s;q)]ς2=Iand b(s;q) = a†(s;q), the appropriate dynamical symmetry algebra is suς(1,1). So the question arises: what are conditions for insuring that such q-commutator takes place? In other words, we have the following Problem 1: To find two operators a(s;q)and b(s;q)and a constant ςsuch that the Hamiltonian H(s;q) = b(s;q)a(s;q)and [a(s;q), b(s;q)]ς=I. For the first part we already have the answer (see Theorem 2.2). The solution of the second one is formulated in the following two theorems. Theorem 3.4 Let H(s;q)be the following difference operator (q-Hamiltonian) H(s;q) = 1 ∇x1(s)A(s)H(s;q)1 A(s).(17) The operators b(s;q) = a↑ α(s;q)and a(s;q) = a↓ α(s;q)given in (12) factorize the Hamiltonian H(s;q)(17) and satisfy the commutation relation [a(s;q), b(s;q)]ς= Λ for a certain real number ςif and only if the following two conditions hold: ∇x(s) ∇x1(s−α)s∇x1(s−1)∇x1(s) ∇x(s−α)∆x(s−α)sσ(s−α)σ(−s−µ+α) σ(s)σ(−s−µ+ 1) =ς, (18) and 1 ∆x(s−α)σ(s−α+ 1) ∇x1(s−α+ 1) +σ(−s−µ+α) ∇x1(s−α)−ς1 ∇x1(s)σ(s) ∇x(s)+σ(−s−µ) ∆x(s)= Λ.(19) Proof: Taking the expression of the operators a↑ α(s) and a↓ α(s), a straightforward calculus shows that a↓ α(s)a↑ α(s) = A1(s)e∂s+A2(s)e−∂s+A3(s)I, where A1(s) = −s∇x1(s+ 1) ∇x1(s) A(s) A(s+ 1)sσ(s+ 1 −α)σ(−s−µ−1 + α) ∆x(s−α)∆x(s+ 1 −α) 1 ∇x1(s+ 1 −α), A2(s) = −s∇x1(s−1) ∇x1(s) A(s) A(s−1)sσ(s−α)σ(−s−µ+α) ∆x(s−1−α)∆x(s−α) 1 ∇x1(s−α), A3(s) = 1 ∆x(s−α)σ(s+ 1 −α) ∇x1(s+ 1 −α)+σ(−s−µ+α) ∇x1(s−α). (20) 7 In the same way, using (10) and (9) we have a↑ α(s)a↓ α(s) = H(s;q) = B1(s)e∂s+B2(s)e−∂s+ B3(s)I, where B1(s) = −1 ∇x1(s) A(s) A(s+ 1) pσ(−s−µ)σ(s+ 1) ∇x(s+ 1) , B2(s) = −1 ∇x1(s) A(s) A(s−1) pσ(−s−µ+ 1)σ(s) ∇x(s), B3(s) = 1 ∇x1(s)σ(s) ∇x(s)+σ(−s−µ) ∆x(s). (21) Consequently, [a↓ α(s), a↑ α(s)]ς=A1(s)−ςB1(s)e∂s+A2(s)−ςB2(s)e−∂s+A3(s)−ςB3(s)I. (22) To eliminate the two terms in the right-hand side of (22), which are proportional to the difference operators exp(±∂s), one must require that A1(s)−ς B1(s) = 0, A2(s)−ς B2(s) = 0.(23) Off hand, it is not evident that one can satisfy both of the relations (23), which only involve the same constant ς. But it is straightforward to verify from (20) and (21) that A1(s)B2(s+ 1) = A2(s+ 1) B1(s), or, equivalently, A1(s) B1(s)=A2(s+ 1) B2(s+ 1). Hence, the requirement that A1(s) = ς B1(s) entails the relation A2(s) = ς B2(s), and vice versa. From (22) it is now evident that the commutator [a↑ α(s), a↓ α(s)]ςis a constant if (23) holds and the factor A3(s)−ς B3(s) is a constant. Thus, the required conditions (18) and (19) immediately follow.  Theorem 3.5 Let (Φn)nthe eigenfunctions of H(s;q)corresponding to the eigenvalues (λn)n and suppose that the problem 1 has a solution for Λ6= 0. Then, the eigenvalues λnof the difference equation (11) are q-linear or q−1-linear functions of n, i.e., λn=C1qn+C3or λn=C2q−n+C3, respectively. Proof: 1In the following we use the notation ς=qγ. Suppose that problem 1 has a solution with Λ6= 0 and λnis not a q-linear (respectively, q−1) function of n. From Proposition 3.2 we know that a↑ α(s;q)Φn(s;q) is and eigenvector of H(s;q) corresponding to the eigenvalue Λ+qγλn. If we denote by Φm(n);qsuch eigenvector where m(n) is a function of n, then we have Λ+qγλn=λm(n). Then using (3) we get λm(n)=C1qm(n)+C2q−m(n)+C3, C1C2=Lq.(24) On the other hand λm(n)= Λ + qγλn=C1qγqn+C2qγq−n+qγC3+ Λ = C0 1qn+C0 2q−n+C0 3.(25) But here, since λm(n)is an eigenvalue of (1), again we have the condition C0 1C0 2=Lq, thus C1C2=Lq=C0 1C0 2=C1C2q2γso q2γ= 1, i.e., γ= 0, or C1C2= 0. In the first case, equating 1For an alternative proof in the case α= 0 see the appendix. 8 (24) and (25), we have that C0 3=C3qγ+ Λ = C3, i.e., Λ = 0 that is a contradiction. Thus C1C2= 0 from where the result easily follows. It is worth noting that the q-linearity of the eigenvalues of H(s;q) is only the necessary condition, i.e., it is not sufficient. So there are cases when the eigenvalues λn(q) are q-linear (for instance, those which correspond to the q-Meixner, the q-Charlier, and the q-Laguerre polynomials with a6=q−1/2), but the corresponding q-Hamiltonians H(s;q) do not admit the factorization in terms of q-commuting operators. This just reflects the fact that an appropriate dynamical algebra is not suq(1,1) and one has to consider a more complicated quadratic algebra AW(3) [21]. The problem of finding an explicit connection between the generators of the algebra AW(3) and the operators a(s;q) and b(s;q), which factorize the q-Hamiltonians for these cases, will be attended in a separate publication. Remark 3.6 A special important case of the non-linear lattice is when x(s) = 1 2(qs+q−s). In this case if we put α=1 2, then the conditions (18) and (19) of Theorem 3.4 becomes sσ(s−1 2)σ(−s+1 2) σ(s)σ(−s+ 1) =ς, and 1 ∇x1(s) σ(s+1 2) ∆x(s)+σ(−s+1 2) ∇x(s)!−ς1 ∇x1(s)σ(s) ∇x(s)+σ(−s) ∆x(s)= Λ. respectively. Moreover, if we put A(s) = 1 then, H(s;q) = (∇x1(s))−1H(s;q)and the α-operators simplify a↓ 1/2(s) = 1 ∇x1(s)e1 2∂spσ(s)−e−1 2∂spσ(−s), a↑ 1/2(s) = 1 ∇x1(s)pσ(s)e−1 2∂s−pσ(−s)e1 2∂s. Now we can formulate the Problem 2: To find two operators a(s;q)and b(s;q)and a constant ςsuch that the Hamiltonian H(s;q) = b(s;q)a(s;q)and [a(s;q), b(s;q)]ς=Iand such that a(s;q)and b(s;q)are the lowering and raising operators, i.e., a(s;q)Φn(s;q) = DnΦn−1(s;q) and b(s;q)Φn(s;q) = UnΦn+1(s;q).(26) Again, without loss of generality, we will change the condition [a(s;q), b(s;q)]ς=Iinto [a(s;q), b(s;q)]ς= Λ and chose Λ = λ1. Also the operators b(s;q) = a↑ α(s;q) and a(s;q) = a↓ α(s;q), given in (12), provide the factorization of H(s;q). If we now apply b(s;q) to the first equation of (26) and use the second one as well as (11), we find that λn=DnUn−1. On the other hand, applying a(s;q) to the second equation in (26) and using the first one as well as (15) we obtain λ1+ςλn=UnDn+1 =λn+1. Thus, using Proposition 2.5 we conclude that λnshould be a ς-linear function, i.e., λnhas the form λn=Aςn+D, where Aand Dare non-vanishing constants. Moreover, using the recurrence λ1+ςλn=λn+1 and Proposition 3.2 we obtain that H(s;q){a(s;q)Φn(s;q)}=λn−1{a(s;q)Φn(s;q)}, H(s;q){b(s;q)Φn(s;q)}=λn+1{b(s;q)Φn(s;q)}, 9 B(x) = c qx(1 −b qx+1), D(x) = (1 −qx)(1 + bc qx). So q-Meixner functions, defined as ΦM n(x;b, c;q) := d−1 n(b, c)cx(bq;q)x (q;q)x(−bcq;q)x1/2 qx(x−1)/4Mn(q−x;b, c;q), are eigenfunctions of a difference “Hamiltonian” HM(x;b, c;q), HM(x;b, c;q)ΦM n(x;b, c;q) = 1−qn 1−qΦM n(x;b, c;q), where HM(x;b, c;q) := 1 1−qhB(x) + D(x)−B1/2(x)e∂xD1/2(x)−D1/2(x)e−∂xB1/2(x)i. The q-Meixner functions satisfy the discrete orthogonality relation ∞ X k=0 ΦM m(k;b, c;q) ΦM n(k;b, c;q) = δmn . One can factorize the “Hamiltonian” HM(x;b, c;q), HM(x;b, c;q) = a↑ M(x;b, c;q)a↓ M(x;b, c;q), by means of the “lowering” and “raising” difference operators a↓ M(x;b, c;q) := 1 √1−qhe∂xD1/2(x)−B1/2(x)i, a↑ M(x;b, c;q) := 1 √1−qhD1/2(x)e−∂x−B1/2(x)i. (29) The difference operators (29) satisfy a q-commutation relation of the form a↓ M(x;b/q, cq;q)a↑ M(x;b/q, cq;q)−qa↑ M(x;b, c;q)a↓ M(x;b, c;q) = I . Their action on the q-Meixner functions is given by a↓ M(x;b, c;q) ΦM n(x;b, c;q) = q1−qn 1−qΦM n−1(x;bq, c/q;q), a↑ M(x;b/q, cq;q) ΦM n(x;b, c;q) = q1−qn+1 1−qΦM n+1(x;b/q, cq;q), (30) that is, they not only lower and raise, respectively, the index n, but alter also the parameters b and c. The formulae (30) are equivalent to the statement that the forward and backward shift operators for the q-Meixner polynomials (28) have the form (see [7], p.95, (3.13.2) and (3.13.8)) (1 −e∂x)Mn(q−x;b, c;q) = 1−qn c(1 −bq)q−xMn−1(q−x;bq, c/q;q), hcqx(1 −bqx)−(1 −qx)(1 + bcqx)e−∂xiMn(q−x;b, c;q) =cqx(1 −b)Mn+1(q−x;b/q, cq;q). It remains only to remind the reader that when the parameter bin (28) vanishes, the qMeixner polynomials Mn(q−x; 0, c;q) coincide with the q-Charlier polynomials ([7], p.112) Cn(q−x;c;q) := 2φ1 q−n, q−x q;−qn+1 c 0 .(31) In this case B(x) = c qxand D(x) = 1 −qx. The appropriate formulae for the q-Charlier polynomials (31) are therefore easy consequences of the corresponding formulae for the q-Meixner polynomials (28) with the vanishing value of the parameter b. 16 4.2 Askey-Wilson cases Using the linearity of the difference equation (1) we can consider, with not loss of generality, the following lattice x(s) = 1 2(qs+q−s), for which µ= 0. Then, σ(s) = Cq−2s 4 Y i=1 (qs−qsi) = q−2s 4 Y i=1 (qs−zi), σ(−s−µ) = Cq2s 4 Y i=1 (q−s−zi). It is well known that the general case when the zeros of ˜σ, namely z1z2z3z46= 0, corresponds to the Askey-Wilson polynomials [7]. We will use the theorem 3.4 to solve the factorization problem for the whole q-Askey tableau [7] and Nikiforov-Uvarov tableau [4, 5]. The Askey-Wilson polynomials on the lattice x(s) = 1 2(qs+q−s) = cos θwhere qs=eiθ, defined by [7] pn(x(s); a, b, c, d|q) = (ab, ac, ad;q)n an4ϕ3 q−n, qn−1abcd, aq−s, aqs ab, ac, ad q;q !, where a=z1,b=z2,c=z3,d=z4and µ= 0. Their orthogonality relation is of the form Z1 −1 ω(x)pn(x;a, b, c, d)pm(x;a, b, c, d)dx =δnmd2 n, where ω(x) = h(x, 1)h(x, −1)h(x, q 1 2)h(x, −q1 2) 2π√1−x2h(x, a)h(x, b)h(x, c)h(x, d), h(x, α) = ∞ Y k=0 [1 −2αxqk+α2q2k], and the norm is given by d2 n=(abcdqn−1, abcdq2n;q)∞ (qn+1, abqn, acqn, adqn, bcqn, bdqn, cdqn;q)∞ . The Askey-Wilson functions can be defined by Φn(s;q) = sω(s)A2(s) d2 n pn(x(s); a, b, c, d), x(s) = cos θ, qs=eiθ. Taking A(s) = p∇x1(s), we have the orthogonality R1 −1Φn(s;q)Φm(s;q)/∇x1(s)dx =δn,m, and H(s;q)Φn(s;q) = λnΦn(s;q), where λn=q(q−n−1)(1 −abcdqn−1), and H(s;q) is given by (9), σ(s) = Cσq−2s(qs−a)(qs−b)(qs−c)(qs−d) and µ= 0. Thus the Hamiltonian, associated with these Askey-Wilson functions, is H(s;q)= −1 k2 q√sin θ pσ(s)σ(−s+ 1) sin(θ+i 2log q)psin(θ+ilog q)e−∂s+pσ(s+ 1)σ(−s) sin(θ−i 2log q)psin(θ−ilog q)e∂s ! +1 k2 qsin θσ(s) sin(θ+i 2log q)+σ(−s) sin(θ−i 2log q)I. If we now use the Remark 3.6, then the first condition of the Theorem 3.4 holds for α=1 2. In fact, a straightforward calculations shows that ∆x1(s+γ) = kq 2q−s−γ(qs+γ+ 1)(qs+γ−1). 17 Hence, the condition (18) has the form (eiθ =qs) q−2α+1  qs−1 2−1 qs−α−1s(qs−1−1)(qs−1) (qs−α−1 2−1)(qs−α+1 2−1)v u u t4 Y i=1 (qs−α−zi)(q−s+α−zi) (qs−zi)(q−s+1 −zi) × qs−1 2+ 1 qs−α+ 1s(qs−1+ 1)(qs+ 1) (qs−α−1 2+ 1)(qs−α+1 2+ 1) =ς. If we look at the expression in the second line, we find that it is a constant if and only if α=1 2, and thus the first condition transforms into sQ4 i=1(qs−1/2−zi)(q−s+1/2−zi) Q4 i=1(qs−zi)(q−s+1 −zi)=ς. (32) Then, the simplest case for which the condition (18) holds is when z1z2=q1 2and z3z4=q1 2 (with the corresponding permutations of the roots). In this case ς= 1. Since σ(s) = σ(−s+1 2), the second condition gives 1 ∇x1(s) σ(s+1 2) ∆x(s)+σ(−s+1 2) ∇x(s)−σ(s) ∇x(s)−σ(−s) ∆x(s)!= 0. Thus, we have a↑ 1/2(s;q)a↓ 1/2(s) = H(s;q) and [a↓ 1/2(s;q), a↑ 1/2(s)]q= 0, where a↓ 1/2(s;q) = e1 2∂ssσ(s) −k2 qsin θsin(θ+i 2log q)−e−1 2∂ssσ(−s) −k2 qsin θsin(θ−i 2log q), a↑ 1/2(s;q) = sσ(s) −k2 qsin θsin(θ+i 2log q)e−1 2∂s−sσ(−s) −k2 qsin θsin(θ−i 2log q)e1 2∂s. Since the α-operators are commuting, this case is not so interesting in applications (e.g., for q-models of the harmonic oscillators). A special case of the Askey-Wilson polynomials are the continuous q-Jacobi polynomials corresponding to the roots a=qα0/2+1/4,b=qα0/2+3/4, c=−qβ0/2+1/4,d=−qβ0/2+3/4[7], then we can solve the problem 1 only in the case when α0=β0=−1/2. The next case is when one of the roots zivanishes. This is the 0-Askey-Wilson polynomials or the continuous dual q-Hahn [7]. In this case the first condition (18) (equivalently (32)) holds only when (z1, z2, z3) = (t, 1 2−t, 1 4), t ∈R. With the above choice of ziit is impossible to fulfill the second condition (19) hence it is impossible to obtain a simple closed dynamical algebra. 4.2.1 Continuous q-Laguerre polynomials Let now consider the case when the Askey-Wilson polynomials have two parameters equal to zero. In order that the condition (18) take place, the other two non-vanishing parameters should satisfy that their product is equal q1 2. Under this condition ς=q−1 2. Then the second condition yields Λ = 4Cσ(√q−1) k2 q. Then, the Askey-Wilson Hamiltonian with two zero roots of σadmits a factorization with a non-trivial dynamical algebra. An example of this family is the continuous 18 q-Laguerre polynomials P(a) n(x|q) [7], x(s) = cos θ, when a=−1/2 for which the Hamiltonian has the form (A(s) = 1) H(s;q)=−Cσ kqsin θ q(qs+1 −1)(qs+1 −q1 2)(q−s−1)(q−s−q1 2) sin(θ+i 2log q)e−∂s+ +q(qs−1)(qs−q1 2)(q1−s−1)(q1−s−q1 2) sin(θ−i 2log q)e∂s + −Cσ kqsin θ (qs−1)(qs−q1 2) sin(θ+i 2log q)+(q−s−1)(q−s−q1 2) sin(θ−i 2log q)!I, and a↓ 1/2(s;q) = √−Cσ kqsin θe1 2∂sq(qs−1)(qs−q1 2)−e−1 2∂sq(q−s−1)(q−s−q1 2), a↑ 1/2(s;q) = √−Cσ kqsin θq(qs−1)(qs−q1 2)e−1 2∂s−q(q−s−1)(q−s−q1 2)e1 2∂s. Then for the functions Φn(x(s)) = s(q1 2;q)n(q, q 1 2;q)∞ω(s) (q;q)n3ϕ2 q−n, q−s, qs q1 2,0q;q  , qs=eiθ, x(s) = cos θ, we have the orthogonality R1 −1Φn(x;q)Φm(x;q)dx =δn,m, and H(s;q)Φn(s;q) = q(q−n−1)Φn(s;q),H(s;q) = a↑ 1/2(s;q)a↓ 1/2(s), and [a↓ 1/2(s;q), a↑ 1/2(s)]q−1/2=4Cσ(√q−1) k2 q , Thus choosing Cσ=−k2 q 4(1−√q)we obtain the relation [a↓ 1/2(s;q), a↑ 1/2(s)]q−1/2=I. The case of Askey-Wilson polynomials with three zero parameters is analogue to the case of one zero parameter and there is not possible to solve the problem 1. An example of this case are the continuous big q-Hermite polynomials [7]. 4.2.2 Continuous q-Hermite polynomials Finally, if one takes the Askey-Wilson polynomials with vanishing parameters a, b, c, d, this gives the continuous q-Hermite polynomials [7]. In this case σ(z) = Cσq2z. Let choose A(s) = p∇x1(s). Taking into account that this family is a special case of the Askey-Wilson polynomials when all parameters a=b=c=d= 0, one directly obtains, by using ς= 1/q, that in this case the Hamiltonian is given by H(s;q)= −1 k2 q√sin θ Cσq sin(θ+i 2log q)psin(θ+ilog q)e−∂s+Cσq sin(θ−i 2log q)psin(θ−ilog q)e∂s! +1 k2 qsin θCσq2s sin(θ+i 2log q)+Cσq−2s sin(θ−i 2log q)I, 19 and the α-operators a↓ 1/2(s;q) = e1 2∂ssCσq2s −k2 qsin θsin(θ+i 2log q)−e−1 2∂ssCσq−2s −k2 qsin θsin(θ−i 2log q) a↑ 1/2(s;q) = sCσq2s −k2 qsin θsin(θ+i 2log q)e−1 2∂s−sCσq−2s −k2 qsin θsin(θ−i 2log q)e1 2∂s , are such that a↑(s;q)a↓(s;q) = H(s;q) and [a↓(s;q), a↑(s;q)]1/q =4Cσ kq . Notice that for getting the normalized commutation relations it is sufficient to choose Cσ=kq/4. Another possible choice is A(s) = 1 [12], hence a straightforward calculus shows that the two conditions in Theorem 3.4 are true if ς=q−1, thus Λ = 4Cσk−1 q. With this choice the orthogonality of the functions Φnis R1 −1Φn(s;q)Φm(s;q)dx =δn,m. In this case, the Hamiltonian is equal to H(s;q) = Cσq k2 qe−∂s sin θsin(θ+i 2ln q)+e∂s sin(θ−i 2ln q) sin θ−4 √q1−1 + q q+q−1−2 cos 2θI, and a↓(s;q) := a↓ 1/2(s) = √−Cσ kqsin θe1 2∂sqs−e−1 2∂sq−s, a↑(s;q) := a↑ 1/2(s) = √−Cσ kqsin θqse−1 2∂s−q−se1 2∂s. In terms of these operators H(s;q) = a↑(s;q)a↓(s;q) and [a↓(s;q), a↑(s;q)]q−1=4Cσ kq . As before we now can choose Cσ=kq/4. This case was first considered in [13], see also [25]. Appendix Here we present a simple proof of the Theorem 3.5 when α= 0. Notice that in this case b(s;q) = a†(s;q), i.e., the operator bis the adjoint of a. Theorem 3.5a: Let {Φn(s;q)}be the eigenfunctions of the operator H(s;q), corresponding to the eigenvalues {λn(q)}. Suppose that the H(s;q) admits the factorization (13). If the operators a(s;q) and b(s;q) in (13) satisfy the q-commutation relation [a(s;q), b(s;q)]q=I, then the eigenvalues λn(q) are q-linear or q−1-linear functions of n, i.e., either λn(q) = C1qn+C3or λn(q) = C2q−n+C3, respectively. Proof: Obviously, the operator H(s;q) is diagonal in the basis consisting of its eigenfunctions Φn(s;q). By hypothesis, this operator admits at the same time the factorization (13) in terms of a(s;q) and b(s;q). But according to proposition 3.2 the function a(s;q)Φn(s;q) is also the eigenfunction of H(s;q), associated with the eigenvalue q−1[λn(q)−1]. Hence the a(s;q) is either the lowering operator or the raising operator. In the former case this means that q−1[λn(q)−1] = λn−1(q) + C , 20 from which it follows that C2= 0 (i.e., the spectrum {λn(q)}is a q-linear one) and C= q−1[(1 −q)C3−1]. In latter case the corresponding relation is q−1[λn(q)−1] = λn+1(q) + C , which holds when C1= 0 (i.e., the spectrum is q−1-linear) and C=q−1[(1 −q)C3−1]. The proof of the theorem is thus complete. Acknowledgments: The research of RAN has been partially supported by the Ministerio de Ciencias y Tecnolog´ıa of Spain under the grant BFM 2003-06335-C03-01, the Junta de Andaluc´ıa under grant FQM-262. 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