Second order q dierence equations solvable by
factorization method
Alina Dobrogowska
Institute of Mathematics, University of Biaªystok
Joint work with A. Odzijewicz
Global Study of Dierential Equations in the Complex Domain
5 September 2013, Warszawa
Alina Dobrogowska
Second order q dierence equations solvable by factorization
The factorization method rst used by Darboux. Later the method
was rediscovered many times, in particular by the founders of
quantum mechanics, while studying the Schrödinger equation.
Infeld and Hull summarized the quantum mechanical applications of
the method.
Alina Dobrogowska
Second order q dierence equations solvable by factorization
We consider the sequence of the eigenvalue problems for the second
order q dierence operators
Hk ψk (x) = Zk (x)∂q Q−1 ∂q + Wk (x)∂q + Vk (x) ψk (x) =
= λk ψk (x),
k ∈ N ∪ {0}.
Zk , Wk , Vk are real-valued functions.
∂q ψ(x) =
ψ(x)−ψ(qx)
(1−q)x
q derivative operator
Q±1 ψ(x) = ψ(q ±1 x)
shift operator
0<q<1
Alina Dobrogowska
Second order q dierence equations solvable by factorization
The operators Hk acting in the Hilbert spaces Hk . By denition
the Hilbert spaces
Hk = L2 ([a, b]q , %k dq x)
consist of the complex valued functions ψ : [a, b]q −→ C dened on
the q interval
[a, b]q = {q n a : n ∈ N ∪ {0}} ∪ {q n b : n ∈ N ∪ {0}}
and square-integrable with respect to the scalar products
hψ|ϕik :=
Rb
a
ψ(x)ϕ(x)%k (x)dq x
dened by Jackson q integral.
Alina Dobrogowska
Second order q dierence equations solvable by factorization
qintegral
0
3
2
a q a q
Ra
0
ψ(x)dq x :=
a q
P∞
n=0 (1
− q)q n aψ(q n a) ,
a
where dq x =
P∞
n=0 (1 − q)xδ(x − q
n a)dx
qintegral on the interval [a,b]
Rb
a
ψ(t)dq t :=
Rb
0
ψ(t)dq t −
Ra
0
ψ(t)dq t ,
Alina Dobrogowska
0
the Fermat measure,
a qb q
a
b
Second order q dierence equations solvable by factorization
• We postulate following recurrence relations between weight
functions:
%k−1 = ηk %k ,
%k−1 = Q (Bk %k ) ,
where ηk , Bk are real valued functions.
• The functions Bk , ηk satisfy the equation
Q (Bk %k ) = ηk %k .
(1)
• Additionally we put a boundary condition
Bk (a)%k (a) = Bk (b)%k (b) = 0 .
After introducing the function:
Ak (x) =
Bk (x) − ηk (x)
(1 − q)x
we obtain from (1) the q Pearson equation
∂q (Bk %k ) = Ak %k .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
We want to factorize the operators Hk
Hk = A∗k Ak + ak ,
for
Ak : Hk → Hk−1
A∗k : Hk−1 → Hk
Ak = ∂q + fk ,
A∗k = (∂q + fk )∗ = −Ak − Bk ∂q Q−1 + fk ηk .
Adjoint operator hA∗k ψk−1 |ϕk ik = hψk−1 |Ak ϕk ik−1 .
We assume the following relations
Ak A∗k + ak = dk A∗k−1 Ak−1 + ak−1
Alina Dobrogowska
∀k ∈ N ∪ {0} .
Second order q dierence equations solvable by factorization
The conditions are equivalent to the following equations
1.
ηk+1 (x) = gk (x)ηk (q −1 x),
2.
ϕk+1 (x) =
3.
ϕ2k (x)ηk (x) −
=
dk+1
ϕk (q −1 x) ,
gk (x)
gk (qx) 2
ϕ (qx)ηk (qx) =
dk+1 k
q 2 dk+1 Bk (qx) − gk (q 2 x)Bk (q 2 x)
+ dk+1 ak − ak+1
(1 − q)2 q 3 x2
where
gk (x) :=
gk (qx)
,
d2k+1
Bk+1 (x)
,
Bk (x)
ϕk (x) := fk (x) +
Alina Dobrogowska
1
.
(1 − q)x
Second order q dierence equations solvable by factorization
Formulas 1 and 2 give the transformation rules equivalent to
ηk (x) = gk−1 (x)gk−2 (qx) . . . g0 (q −k+1 x)η0 (q −k x) ,
Bk (x) = gk−1 (x)gk−2 (x) . . . g0 (x)B0 (x) ,
ϕk (x) =
dk ...d1
ϕ (q −k x)
gk−1 (x)...g0 (q −k+1 x) 0
Alina Dobrogowska
.
Second order q dierence equations solvable by factorization
Let us dene the function αk (x) := ϕ2k (x)ηk (x), and insert it into
equations 3 to obtain
α0 (x)−
gk (q k+1 x))Gk (x)
gk (q k+1 x) Gk (x)
α0 (qx) =
(dk+1 ak − ak+1 +
dk+1 Gk (qx)
d2k+1
q 2 dk+1 gk−1 (q k+1 x) . . . g0 (q k+1 x)B0 (q k+1 x) − gk (q k+2 x) . . . g0 (q k+2 x)B0 (q k+2 x
(1 − q)2 q 2k+3 x2
for k ∈ N ∪ {0}, where Gk (x) :=
G0 (x) ≡ 1.
Alina Dobrogowska
gk−1 (q k x)...g0 (qx)
(dk ...d1 )2
for k ∈ N,
Second order q dierence equations solvable by factorization
This is innite sequence of equations on one function α0 . We
postulate that it reduces in fact to one equation. This requirement
is equivalent to the following
1.
gk (q k+1 x) Gk (x)
g0 (qx)
=
,
dk+1 Gk (qx)
d1
2.
q 2 dk+1 gk−1 (q k+1 x) . . . g0 (q k+1 x)B0 (q k+1 x) − qk (q k+2 x) . . . g0 (q k+2 x)B0 (q k+
(1 − q)2 q 2k+3 x2
gk (q k+1 x)Gk (x)
d2k+1
2
q d1 B0 (qx) − g0 (q 2 x)B0 (q 2 x)
g0 (qx)
=
+ d1 a0 − a1
2
3
2
(1 − q) q x
d21
+dk+1 ak − ak+1 )
Alina Dobrogowska
k∈N .
Second order q dierence equations solvable by factorization
The rst condition can be reduced to
gk (x) =
dk+1
g0 (x) .
d1
From this we infer that the equation 2 are equivalent to
q 2 d1 1 − g0 (qx)Q 1 − (qd1 )−2k g0 (q k x) . . . g0 (qx)g0k (q k x)Qk B0 (x) =
= (1−q)2 qx2
dk+1 ak − ak+1 −k+1
g0 (q k x) . . . g0 (qx) − d1 a0 + a1
d1
dk+1 . . . d1
,
k ∈ N. We have the following series of equations for the function
B0 . We make the assumption that the series reduces to one
equation.
Alina Dobrogowska
Second order q dierence equations solvable by factorization
This requirement is equivalent to the following
1.
g0 (x) = d1 q γ
γ ∈ R,
B0 (x) = x2 b2 + b1 x−γ + b0 x−2γ ,
[γk]
−γk
+
3.
ak+1 = dk+1 . . . d1 q
−a0
[γ]
a1 [γ(k + 1)]
+
− qb2 [γk][γ(k + 1) ,
k∈N
d1
[γ]
2.
Substituting B0 to (*) we obtain
α0 (x) =
q γ+1 b2
q γ (d1 a0 − a1 )
q 1−γ b0 −2γ
−γ
+
+
hx
+
x
,
(1 − q)2
(1 − q γ )d1
(1 − q)2
Alina Dobrogowska
Second order q dierence equations solvable by factorization
Finally we obtain the solutions
Bk (x) = q kγ dk . . . d1 B0 (x) ,
ηk (x) = q kγ dk . . . d1 η0 (q −k x) ,
ϕk (x) = q kγ ϕ0 (q −k x) ,
αk (x) = q −kγ dk . . . d1 α0 (q −k x) .
Ak (x) = q kγ dk . . . d1 q −k A0 (q −k x) + [−2k]b2 x+
+[k(γ − 2)]b1 x−γ+1 + [2k(γ − 1)]b0 x−2γ+1 ,
fk (x) = q −kγ f0 (q −k x) −
1 − q k(1−γ)
,
(1 − q)x
γk(k−1)
q− 2
%0 (q −k x)
%k (x) =
,
Q
dk d2k−1 . . . dk1 0n=−k+1 B0 (q n x)
Alina Dobrogowska
Second order q dierence equations solvable by factorization
where
s
f0 (x) =
%0 (x) =
α0 (x)
1
−
,
B0 (x) − (1 − q)xA0 (x) (1 − q)x
B0 (qx)
%0 (qx).
B0 (x) − (1 − q)xA0 (x)
Alina Dobrogowska
Second order q dierence equations solvable by factorization
The creation and annihilation operators are in this case given by
s
1
α0 (q −k x)
−γk
Ak = ∂q −
+q
,
(1 − q)x
η0 (q −k x)
A∗k
= dk . . . d1 −q (b2 x + b1 x
γk
2
2−γ
2−2γ
+ b0 x
q
−k
−k
+ α0 (q x)η0 (q x)
Alina Dobrogowska
−1
) ∂q Q
1
+
(1 − q)x
+
Second order q dierence equations solvable by factorization
Hamiltonian takes the form
Hk = dk . . . d1
s
−(1 − q)q −1 x3 B0 (x)
α0 (q −(k+1) x)
∂q Q−1 ∂q +
η0 (q −(k+1) x)
s
q
α0 (q −(k+1) x)
+
α0 (q −k x)η0 (q −k x
η0 (q −(k+1) x)
s
!
b2 + b1 x−γ + b0 x−2γ
α0 (q −(k+1) x)
+
q − (1 − q)x
+
(1 − q)2
η0 (q −(k+1) x)
q
1
−γk
−k
+q
α0 (q x) −
η0 (q −k x)α0 (q −k x)+
(1 − q)x
[γ(k − 1)] a1 [γk]
−γ(k−1)
−
+ qb2 [γ(k − 1)][γk]
,
−q
a0
[γ]
d1 [γ]
+ −q −1 x2 (b2 + b1 x−γ + b0 x−2γ )
Alina Dobrogowska
Second order q dierence equations solvable by factorization
Let us now present the limit behaviour of the formulas obtained
above when the parameter q tends to 1. It is easy to see that the
set [a, b]q becomes the interval [a, b] in the limit q → 1 and the
Rb
scalar product turns to be hψ|ϕik = a ψ(x)ϕ(x)%k (x)dx , where
the weight function %k (x) satises Pearson equation
d
dx (%k Bk ) = %k Ak . For q → 1 the operator Q goes to the identity
d
operator and ∂q −−−→ dx
. In the limiting case the annihilation and
q→1
creation operators are of the form
d
+ fk ,
dx
d
∗
Ak = Bk −
+ fk − Ak
dx
Ak =
and the operators Hk are given by
Hk = −Bk
d
d2
− Ak
+ (fk2 − fk0 )Bk − fk Ak + ak .
2
dx
dx
Alina Dobrogowska
Second order q dierence equations solvable by factorization
The recurrence transformations
for q → 1 tend to Bk+1 = dk+1 Bk ,
d
Ak+1 = dk+1 Ak − dx Bk . The sequence of q -dierence
equations tends to the sequence of nonlinear dierential equations
2
0
Bk (fk+1
−fk2 +fk+1
+fk0 )−Ak (fk+1 −fk )+2Bk0 fk+1 −A0k +Bk00 = ak −
ak+1
,
dk+1
k ∈ N ∪ {0}. The equation for Bk (x) ≡ 1 and Ak (x) ≡ 0 was
considered in many papers, but nevertheless for these
dierentialdierence equations there is no complete theory. One of
the methods for solvingPis to look for the solutions in the form of
innite series fk (x) = i∈Z f˜i (x)k i and obtain in this way the
conditions on the function f˜i (x). The case of solutions given by the
nite series were consider by Infeld and Hull. The classication of
all factorisable onedimensional problems is still an open question.
Alina Dobrogowska
Second order q dierence equations solvable by factorization
Eigenproblem for the chain of operators
Hk ψkn = λnk ψkn
∗
Hk = A∗k Ak + ak = d−1
k+1 Ak+1 Ak+1 + ak+1 .
The ground states are the solutions of the following equations
Ak ψk0 = 0
with the eigenvalues
λ0k = ak .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
The factorization method gives us the following eigenfunctions of
Hk
0
ψkn (x) := A∗k . . . A∗k−n+1 ψk−n
(x),
n = 1, . . . , k ,
with the eigenvalues
λnk = dk dk−1 . . . dk−n+1 ak−n .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
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Alina Dobrogowska
•
/
k
Second order q dierence equations solvable by factorization
Example: class of q Hahn polynomials
fk (x) ≡ 0, k ∈ N ∪ {0}
∧
dk = q −1
These conditions can be reduced to
B0 (x) = b2 x2 + b1 x + b0 ,
a1
b1
A0 (x) = (1 + q)b2 − q(a0 − ) x +
− (1 − q)h .
d1
1−q
We have γ = 1,
Bk (x)
Ak (x)
=B0 (x) ,
1 − Q−k
−k
−k
= q A0 (q x) +
B0 (x) = aek x + bek ,
(1 − q)x
where
aek = −q 1−2k (q[2(k − 1)]b2 + (a0 − qa1 )) ,
b1
− (1 − q)q −k h .
bek =
1−q
Alina Dobrogowska
Second order q dierence equations solvable by factorization
In this case the annihilation and creation operators are given by
Ak = ∂q ,
A∗k = −(b2 x2 + b1 x + b0 )∂q Q−1 − aek x − bek
and the Hamiltonian
Hk = −B0 (x)∂q Q−1 ∂q − Ak (x)∂q +
+q −2k+1 (−a0 [k − 1] + qa1 [k] − qb2 [k − 1][k]) .
I
I
I
Let %0 (x) be the solution of the q Pearson equation
%0 (q −k x)
∂q (B0 %0 ) = A0 %0 . Then %k (x) = B0 (q−k+1
is the
x)...B0 (x)
solution of q Pearson equation with functions Ak (x), Bk (x).
%0 and %k satisfy the same boundary conditions.
If {Pkn (x)}∞
n=0 is the orthogonal polynomial system
corresponding to the Hamiltonian Hk then {∂q Pkn (x)}∞
n=0 is
the orthogonal polynomial system corresponding to Hk−1 .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
Considered equation has the form
B0 ∂q Q−1 + Ak ∂q Pkn = λnk Pkn
and we obtain the solutions of the Hahn equation
Pk0 = 1 ,
Pkn (x) = A∗k A∗k−1 . . . A∗k−n+1 1
for eigenvalues
n = 1, 2, ...k ,
λ0k = 0
λnk = aek [n] + b2 [n][n − 1]q −(n−1) .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
Example
%k (x) = const This is equivalent to these conditions
γ=1
∧
b2 = b1 = 0 ,
γ=2
∧
b2 = b0 = 0 .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
Example 2
q deformation of the harmonic oscillator
γ=1
dk = q −1
b0 = 1
Bk (x) = 1 ,
Ak (x) = 0 ,
fk (x) = q −k f0 (q −k x) ,
%k = %0 = 1 ,
where
f0 (x) =
s
q 2 (q −1 a0 − a1 )
1
1
1
1
+h +
−
.
(1 − q)
x (1 − q)2 x2 (1 − q)x
Alina Dobrogowska
Second order q dierence equations solvable by factorization
In this case the annihilation and creation operators are given by
Ak = ∂q + q −k f0 (q −k x) ,
A∗k = −∂q Q−1 + q −k f0 (q −k x) .
and Hamiltonian has the form
Hk = − 1 + (1 − q)q −k−1 xf0 (q −k−1 x) ∂q Q−1 ∂q +
+q −k f0 (q −k x) − q −1 f0 (q −k−1 x) ∂q +
−q −k ∂q (f0 (q −k−1 x))+q −2k f02 (q −k x)+q −2k a0 + (q 2 a1 − a0 )[k] .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
Solutions of the equations
Ak ψk0 = 0
have the form
1.
ψk0 (x) = r
Ck0
q −k x
x1 ; q ∞
q −k x
x2 ; q ∞
q −1 a0 − a1 6= 0
for
where x1 i x2 are the roots of the polynomial
(1 − q)q 2 (q −1 a0 − a1 )x2 + (1 − q)2 hx + 1 = 0 .
2.
ψk0 (x) = r
Ck0
−(1 − q)2 bh0 q −k x; q
Alina Dobrogowska
for
q −1 a0 = a1 h 6= 0 .
∞
Second order q dierence equations solvable by factorization
The operators Ak , A∗k , Q and Q−1 satisfy the relations
qA∗k Q−1 = Q−1 A∗k−1 ,
A∗k Q = qQA∗k+1 ,
qAk Q−1 = Q−1 Ak−1 ,
Ak Q = qQAk+1 .
It easy to see that the operator Q−1 acts as follows
0
C1
−1
0Q
C0
0
C2
−1
0Q
C1
0
Ck
Q−1
C0
k−1
0
Ck+1
Q−1
C0
0 k
/ ψ0
/ ...
/ ... .
/ψ
ψ00
1
k
The functions ψk0 are eigenvectors of the Hamiltonians Hk with the
eigenvalues
λ0k = ak = q −2k a0 + (q 2 a1 − a0 )[k] .
Similarly it is easy to show that the functions
ψkn (x) = Q−k ψ0n (x)
are eigenvectors of Hk with
λnk = q −2k λn0 + (q 2 a1 − a0 )[k] .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
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 −1
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 −1



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Q−1
Q
Alina Dobrogowska
/
k
Second order q dierence equations solvable by factorization
The functions
ψkn (x) = q
1
(a0 − qa1
)n n
q
!q n(n−1)+k
Qn−k A∗n . . . A∗1 ψ00 (x) ,
for k ∈ N ∪ {0} and n ∈ N ∪ {0}, are the eigenvectors of
Hamiltonians corresponding to the eigenvalues
λnk = q −2k+n a0 + (q 2 a1 − a0 )[k − n] .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
It is easy to show that in the limit q → 1 this case gives us the
harmonic oscillator
Hk = −
d2
(a0 − a1 )2 2 a1 + a0
+
x +
+ (a1 − a0 )k
dx2
4
2
with eigenvectors
d
a0 − a1 n − a0 −a1 x2
4
ψkn (x) = −
+
x e
for n ∈ N ∪ {0}
dx
2
corresponding to the eigenvalues
λnk = a0 + (a0 − a1 )(n − k) .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
We have showed in the paper: A. Dobrogowska, A. Odzijewicz
”Second order q dierence equations solvable by factorization
method”, J. Comput. Appl. Math. 193 (2006) 319-346, that
Ak =
q −kγ 1−γ
x ∂q + fk ,
[γ]
q k(γ−1) γ−1
x Bk ∂q Q−1 + Bk fk −
[γ]
− Ak 1 + (1 − q γ )q kγ xγ fk ,
1−k
γ [k − 1][k]
a0 [k − 1] − a1 [k] + q b2
ak = − q
,
[γ]2
A∗k = −
Alina Dobrogowska
Second order q dierence equations solvable by factorization
where
Bk (x) = q 2kγ−k x2(γ−1) B(q k x),
q kγ−k γ−2 k
2k(1−γ)
Ak (x) =
x
B(q
x)
−
q
B(x)
,
1 − qγ
s
γ−1
D(qx)
1
q −k+ 2
fk (x) =
−
,
γ
γ
γ
(1 − q )x
B(x)
(1 − q )q kγ xγ
and
B(x) =b2 x2 + b1 x + b0 ,
D(x) = b2 + (1 − q γ )[γ]q −γ (a0 − a1 ) x2 +
(b1 + (1 − q γ )c) x + b0 ,
[k] =
1 − qk
= 1 + q + . . . q k−1 .
1−q
Alina Dobrogowska
Second order q dierence equations solvable by factorization
Having in mind the factorization property
A∗k Ak + ak = Q−1 Ak+1 A∗k+1 Q + ak
we can look for the solutions of the eigenvalue problem in the form
ψkn (x) = Pkn (x)ψk0 (x)
under the condition
λnk = ak−n .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
On
a−2 • P02 ψ00
a−1 • P12 ψ10
a0 • P22 ψ20
a−1 • P01 ψ00
a0 • P11 ψ10
a1 • P21 ψ20
a0 • P00 ψ00
a1 • P10 ψ10
a2 • P20 ψ20
O
O
O
O
Alina Dobrogowska
O
O
/
k
Second order q dierence equations solvable by factorization
We obtain the second order q dierence equation (Hahn equation)
−q −1 D(qx)Pkn (qx) − B(q k x)Pkn (q −1 x)+
+ q −1 D(qx) + B(q k x) Pkn (x) =
= (1 − q n ) qb2 + (1 − q γ )[γ]q −γ+1 (a0 − a1 )−
−q 2k−n b2 x2 Pkn (x)
solutions to which could be expressed in terms of the basics
hypergeometric series
∞
X
(a1 ; q)k (a2 ; q)k (a3 ; q)k k
a1 , a2 , a3 q; x =
x
3 Φ2
d1 , d 2
(d1 ; q)k (d2 ; q)k (q; q)k
k=0
(a; q)k = (1 − a)(1 − qa) . . . (1 − q k−1 a)
(a; q)∞ = (1 − a)(1 − qa)(1 − q 2 a) . . .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
∂q
q −kγ 1−γ
B k %k
[γ] x
= Ak %k q Pearson equation
x1−2γ Bk %k |ψk0 |2 |ba = 0
boundary conditions
2
B(x) =b2 x + b1 x + b0 = b0
x
1−
x1
x
1−
x2
,
D(x) = b2 + (1 − q γ )[γ]q −γ (a0 − a1 ) x2 +
(b1 + (1 − q γ )c) x + b0 =
x
x
1−
.
= b0 1 −
y1
y2
Alina Dobrogowska
Second order q dierence equations solvable by factorization
The case of the Big q Jacobi orthogonal
polynomials
( b2 6= 0, b0 6= 0 and b2 + (1 − q γ )[γ]q −γ (a0 − a1 ) 6= 0)
The solutions are
v
u x x u
;q
;q
0
(k+ 21 )(γ−1) u x1 ∞ x2 ∞
t
ψk (x) = Cx
,
qx
qx
y1 ; q
y2 ; q
∞
Pkn (x) = 3 φ2
q −n , q −2k+n+1 xy1 x1 2 y2 , yq1 x
q −k+1 xy12 , q −k+1 xy11
∞
!
q; q
and the eigenvalue is
λnk = a0 [n − k + 1] − qa1 [n − k] +
Alina Dobrogowska
q γ b2
[n − k + 1][k − n] ,
[γ]2
Second order q dierence equations solvable by factorization
the weight function
%k (x) = x(1−γ)(2k+1)
x
x
;
q
;
q
x1
x2
k+1
q interval
,
k+1
[a, b]q = [q −k x1 , q −k x2 ]q .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
In the limit q → 1 (we have assumed that b2 and b1 do not depend
on the parameter q ):
1
1
ψk0 (x) = Cx(γ−1)(k+ 2 ) (x − x1 )
× (x − x2 )
1−γ 2 (a0 −a1 )
+
2
1−γ 2 (a0 −a1 )
−
2
γ 2 b1 (a0 −a1 )
−γc
2b2
2
b2
−4b
b
2 0
1
γ 2 b1 (a0 −a1 )
−γc
2b2
−4b
b
2
b2
2 0
1
√
×
√
,
the Jacobi orthogonal polynomials
(αk + 1)n
Pkn (x) = Pn(αk ,βk ) (y) =
×
n!
−n, n + αk + βk + 1 1 − y
×2 F1
,
2
αk + 1
1
where
γ 2 (a0 − a1 )
γ(γ(a0 − a1 )x1 + c)
−k−
,
b2
b2 (x2 − x1 )
γ(γ(a0 − a1 )x1 + c)
βk = −1 +
.
b2 (x2 − x1 )
αk = −
Alina Dobrogowska
Second order q dierence equations solvable by factorization
After the transformation given by
s
p
∆
b1
−1
x=
cosh
γ
b
(z
−
c)
−
2
2b2
4b22
or
s
x=
p
∆
b1
sin γ −1 |b2 |z −
,
2
2b2
4b2
we obtain the Schrödinger equation with the RosenMorse II
potential (for b2 > 0)
p
p
Vk (z) = D1 coth γ −1 b2 (z − c) cosech2 γ −1 b2 (z − c)+
p
+D2 cosech2 γ −1 b2 (z − c) + D3
or the Eckart II potential (for b2 < 0)
p
p
Vk (z) = D1 tan γ −1 |b2 |z sech γ −1 |b2 |z+
p
+D2 sech2 γ −1 |b2 |z + D3 ,
where D1 and D2 depend on the parameters b2 , b1 , b0 , a0 , a1 , c, γ .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
The case of the Big q Laguerre orthogonal
polynomials
(b2 6= 0, b0 6= 0, b2 + (1 − q γ )[γ]q −γ (a0 − a1 ) = 0 and
b1 + (1 − q γ )c 6= 0)
The solutions are
v
u
x
x
u
;
q
;
q
x1
1
u
∞ x2 ∞ ψk0 (x) = Cx(k+ 2 )(γ−1) t − bb10 + (1 − q γ ) bc0 qx; q
,
∞

Pkn (x) =3 φ2 
−
b1
b0
q −n , 0, − bb10 + (1 − q γ ) bc0 qx
+ (1 − q γ ) bc0 q 1−k x1 , − bb01 + (1 − q γ ) bc0 q 1−k x2

q; q 
and the eigenvalue is
λnk = q k−n (a0 [n − k + 1] − a1 [n − k]) ,
Alina Dobrogowska
Second order q dierence equations solvable by factorization
the weight function
%k (x) = x(1−γ)(2k+1)
x
x
;
q
;
q
x1
x2
k+1
q interval
,
k+1
[a, b]q = [q −k x1 , q −k x2 ]q ,
[a, b]q = [q −k x1 , q −k x2 ]q .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
In the limit q → 1 (we have assumed that b1 does not depend on
the parameter q ):
1
ψk0 (x) = Cx
γ−1
−k
2
x+
b0
b1
b1 −c+γ
2 (a −a ) b0
0
1 b
1
2b1
γ 2 (a0 − a1 )
× exp −
x ,
2b1
the Laguerre polynomials
(αk + 1)n
−n
1 n
(αk )
Pk (x) = Ln (y) =
1 F1
αk + 1
n!
where
×
y ,
γ 2 (a0 − a1 )b0
− γc + k ,
b21
x(2k+1)(1−γ)
%1k (x) =
,
(x + Bb10 )k+1
αk =
[a, b]1 =[−
Alina Dobrogowska
b0
, ∞] ,
b1
Second order q dierence equations solvable by factorization
After the transformation given by
x=
γ −2 b1 2 b0
z −
,
4
b1
we obtain the Schrödinger equation with the threedimensional
isotropic harmonic oscillator potential
Vk (z) = D1 z 2 +
D2
+ D3 ,
z2
where D1 and D2 depend on the parameters b1 , b0 , a0 , a1 , c, γ .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
The case of the AlSalamCarlitz II orthogonal
polynomials
(b2 6= 0, b0 6= 0, b2 + (1 − q γ )[γ]q −γ (a0 − a1 ) = 0 and
b1 + (1 − q γ )c = 0)
The solutions are
s
x
x
0
(k+ 12 )(γ−1)
ψk (x) = Cx
;q
;q
,
x1
∞ x2
∞
!
k −1 n n
k+1
q
−n
x
q
q , x2
2
1
q;
Pkn (x) = −
q 2 φ1
x
x2
x1
0
and the eigenvalue is
λk = q k−n (a0 [n − k + 1] − a1 [n − k]) ,
the weight function
%k (x) = x(1−γ)(2k+1)
x
x
;
q
;
q
x1
x2
k+1
q interval
Alina Dobrogowska
,
k+1
Second order q dierence equations solvable by factorization
In the limit q → 1
1
ψk0 (x)
= Cx
γ−1
−k
2
γ 2 (a0 − a1 ) 2
γc
exp −
x −
x ,
b0
2b0
the Hermite polynomials
1
Pkn (x)
n
= Hn (y) = (2y) 2 F0
− n2 , − n−1
2
− 1
y2 ,
−
%1k (x) =x(2k+1)(1−γ) ,
[a, b]1 =[−∞, ∞] .
After the transformation given by
x = γ −1
p
b0 z
we obtain the Schrödinger equation with the harmonic oscillator
potential
Vk (z) = D1 z 2 + D2 ,
where D1 and D2 depend on the parameters b0 , a0 , a1 , c, γ .
Alina Dobrogowska
Second order q dierence equations solvable by factorization
References:
Dobrogowska, A., Goli«ski, T., Odzijewicz, A, Change of
variables in factorization method for second order functional
equations, Czech. J. Phys. 54 (2004), no. 11, 1257-1263.
Dobrogowska, A., Odzijewicz, A., Second order qdierence
equations solvable by factorization method, J. Comp. Appl.
Math. 193 (2006), 319346.
Dobrogowska, A., Odzijewicz, A., Solutions of the qdeformed
Schrrödinger equation for special potentials, J. Phys. A: Math.
Theor. 40 (2007), 20232036.
Dobrogowska, A., The q-deformation of the Morse potential,
Appl. Math. Lett. 7 (2013), 769773.
Alina Dobrogowska
Second order q dierence equations solvable by factorization