EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 4, Article Number 6733
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Matrix Representations of the Two-Parameter
Deformed Oscillator Algebra

Latif A-M. Hanna1,∗, Shoukry S. Hassan2, Maali Almutairi1
1 Department of Mathematics, Faculty of Science, Kuwait University, P.O. Box 5969,
Safat 13060, Kuwait City, KUWAIT
2 Independent Researcher, 237 Banafseg 7, New Cairo, Egypt

Abstract. Faithful matrix representations are presented for Polynomially (p, q)-deformed Lie al-
gebra lp,q : [K0,K+]p,q = rK+, [K−,K0]p,q = rK−,[K+,K−]p,q = F (K0), where F is a real
polynomial function, and p, q, r ∈ R∗, (R∗ is the set of nonzero real numbers). Conditions are
set for F and the (p, q)-parameters, where the operators K+,K−, and K0, satisfy certain physical
properties.
2020 Mathematics Subject Classifications: 17B10, 17B81, 35Q40, 81V80
Key Words and Phrases: Faithful matrix representation of Lie algebras, deformed Lie algebra,
(p, q)-algebraic deformation

1. Introduction

In many physical contexts, such as quantum optical systems, non-linear quantum
Hamiltonians of order higher than quadratic or bilinear forms appear in modelling and
investigations of many phenomena. Few examples are: many-body systems and multi-
photon processes [1], [2], particles that interpolate between Bosons and Fermions [3], [4],
[5], and quantum deformed (q-deformed) oscillators. Such higher orders of non-linear
Hamiltonian models are usually associated with two types of non-linear q-deformed al-
gebra, namely, the q-deformed Lie brackets and the polynomially deformed Lie algebra
SUpd (2), [6].

Generally speaking, within the quantum mechanical context, generalization of the har-
monic oscillator (HO) algebraic commutation relations, called q-deformed HO, is due to
basically the mathematical non-linearity/non-ideal nature of the concerned quantum com-
plex system [[7], and refs. therein]. Two particular physical applications of the deformation
algebra, namely: (i) The one q-deformation parameter is an indicator of information of

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6733

Email addresses: latif.hanna@ku.edu.kw (L. A-M. Hanna),
shoukryhassan@hotmail.com (S. S. Hassan), maali.almutairi@ku.edu.kw (M. Almutairi)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



L. A-M. Hanna, S. S. Hassan, M. Almutairi / Eur. J. Pure Appl. Math, 18 (4) (2025), 6733 2 of 12

quantum impurities in many-particle interactions in a condensed matter system [8], (ii)
The two (p, q)-deformation parameters of the bosonic Fibonacci oscillator algebra [9] mimic
the defects and impurities in crystalline lattice structure.

Investigations of appropriate time evolution operators for such nonlinear Hamiltonian
systems are generally complicated to handle. The alternative Lie algebraic decomposition
approach and its faithful matrix representation (once found!) for the generators of such
deformed Lie algebra is relatively less tedious than dealing with the direct method of
solving Schrödinger’s wave equations, e.g., [10].

On the other hand, the q-deformed Virasoro algebras [11] are ideal physical applications
of quantum groups [12], which are non-commutative and co-commutative Hopf algebras.
A two-parameter quantum deformation of Lie super algebras in the non-standard simple
root system with two odd simple roots is examined in [13].

In the present work, we search for faithful matrix representations of the generators of
two-parameter (p, q)-deformed algebraic structure, namely, the q- and the polynomially-
deformed Lie algebra, associated with the generalized algebra, lp,q, mentioned next in
Section 2.

2. Preliminaries

In [12], the (p, q)-deformed Lie bracket is given as follows.

Definition 1. Let p and q be real numbers. If X and Y are square matrices of order n,
then the (p, q)-deformed Lie bracket of X and Y , namely [X,Y ]p,q, is defined as,

[X,Y ]p,q = pXY − qY X. (1)

For the two particular cases, namely, p = 1, q = 0 and p = 0, q = −1, we have the
usual matrix multiplication of X and Y . So we always consider that here, p and q are
supposed to be nonzero real numbers, i.e., pq 6= 0. For instance, in [3], the model of
Fermion Oscillators has p = 1 and 0 ≤ q ≤ 1. It should be noted that the case when
p = 1, q = 1 is the ordinary Lie bracket. Thus, we may write [X,Y ]1,1 as [X,Y ].

Faithful matrix representations of the least degree of the Lie algebra l were considered
in [10]-[14], [1]-[15] where,

l : [K0,K±] = ±rK± and [K+,K−] = F (K0) , (2)

with F is a real polynomial function and r ∈ R∗ (R∗, is the set of nonzero real numbers),
subject to the physical properties, namely, K0 is a real diagonal operator and K− = K†

+,
(† is for Hermitian conjugation). The nonzero parameter r is a proportionality factor
relating the HO operators K± to the corresponding atomic spin-up and -down operators
S± (e.g., [16])

In the present work, we consider the Hamiltonian H of two coupled harmonic oscillators
in an optical cavity, representing ’optical 2-level atom’ in the form [16] (we take the reduced
Planck constant h= 1),



L. A-M. Hanna, S. S. Hassan, M. Almutairi / Eur. J. Pure Appl. Math, 18 (4) (2025), 6733 3 of 12

H = ωK0 + λ (t) (K+ +K−) = H†. (3)

Note that, the total Hamiltonian operator H in (3) represents the total quantum
energy operator of the considered system (coupled quantized optical atoms). For physical
requirements, H must be a Hermitian operator, and so, its observed energy eigenvalues
must be real, and in turn, its constituent operators K0, a real diagonal operator, and
K− = K†

+ are two Hermitian conjugate operators.
Here, we examine the faithful matrix representations of the generalized Lie algebra lp,q

(cf [13]),
[K0,K+]p,q = rK+, (4)

[K−,K0]p,q = rK−, (5)

[K+,K−]p,q = F (K0) , (6)

subject to the physical properties mentioned below (3). Note that, for p = q = 1, equations
(4)-(6) are reduced to the Lie algebra l1,1 = l in (2), [14], [15].

3. Basic Properties of the (p, q)-deformed Lie Bracket

Here, we list some properties of the (p, q)-deformed Lie bracket in the following theo-
rem.

Theorem 1. Let X,Y and Z be n× n matrices, and p, q ∈ R∗, and α, β ∈ R. Then

1. tr
(
[X,Y ]p,q

)
= tr

(
[Y,X]p,q

)
= (p− q) tr (XY ) = (p− q) tr (Y X) .

2. [X,Y ]†p,q =
[
Y †, X†]

p,q
.

3.
([

X,X†]
p,q

)
ij
=

(
[X,X†]p,q

)
ji

for i 6= j.

4. [X,Y + Z]p,q = [X,Y ]p,q + [X,Z]p,q .

5. [X + Y, Z]p,q = [X,Z]p,q + [Y, Z]p,q .

6. [αX, βY ]p,q = αβ [X,Y ]p,q .

7.
[
X, [Y, Z]p,q

]
p,q

= p2XY Z − pqY ZX − pqXZY + q2ZY X.

8.
[
[X,Y ]p,q , Z

]
p,q

= p2XY Z − pqY XZ − pqZXY + q2ZY X.

9.
[
X, [Y, Z]p,q

]
p,q

+
[
Y, [Z,X]p,q

]
p,q

+
[
Z, [X,Y ]p,q

]
p,q

= (p− q) [p (XY Z + Y ZX + ZXY )− q (XZY + Y XZ + ZY X)] .



L. A-M. Hanna, S. S. Hassan, M. Almutairi / Eur. J. Pure Appl. Math, 18 (4) (2025), 6733 4 of 12

Proof. The proof of parts (1) - (6) is simple and deduced directly from the definition
in (1). So, we prove parts (7) - (9).

For the proof of part (7), we have[
X, [Y, Z]p,q

]
p,q

= [X, (pY Z − qZY )]p,q

= pX (pY Z − qZY )− q (pY Z − qZY )X
= p2XY Z − pqXZY − pqY ZX + q2ZY X.
Similarly, for the proof of part (8), we have[
[X,Y ]p,q , Z

]
p,q

= [(pXY − qY X) , Z]p,q

= p (pXY − qY X)Z − qZ (pXY − qY X)
= p2XY Z − pqY XZ − pqZXY + q2ZY X.
(iii) Now, for part (9), we have[
X, [Y, Z]p,q

]
p,q

+
[
Y, [Z,X]p,q

]
p,q

+
[
Z, [X,Y ]p,q

]
p,q

= [X, (pY Z − qZY )]p,q + [Y, (pZX − qXZ)]p,q + [Z, (pXY − qY X)]p,q
= {pX (pY Z − qZY )− q (pY Z − qZY )X}+{pY (pZX − qXZ)− q (pZX − qXZ)Y }+

{pZ (pXY − qY X)− q (pXY − qY X)Z}
= p2 (XY Z + Y ZX + ZXY )−pq (XZY + Y ZX + Y XZ + ZXY + ZY X +XY Z)+

q2 (ZY X +XZY + Y XZ)
= p2 (XY Z + Y ZX + ZXY )−pq (XY Z + Y ZX + ZXY +XZY + Y XZ + ZY X)+

q2 (XZY + Y XZ + ZY X)
= p2 (XY Z + Y ZX + ZXY )−pq {(XY Z + Y ZX + ZXY ) + (XZY + Y XZ + ZY X)}+

q2 (XZY + Y XZ + ZY X)
= p2 (XY Z + Y ZX + ZXY )−pq (XY Z + Y ZX + ZXY )−pq (XZY + Y XZ + ZY X)+

q2 (XZY + Y XZ + ZY X)
=

(
p2 − pq

)
(XY Z + Y ZX + ZXY )−

(
pq − q2

)
(XZY + Y XZ + ZY X)

= p (p− q) (XY Z + Y ZX + ZXY )− q (p− q) (XZY + Y XZ + ZY X)
= (p− q) [p (XY Z + Y ZX + ZXY )− q (XZY + Y XZ + ZY X)] .

Part (9) of Theorem 1 is the Jacobi identity, where its right-hand side should be zero
for p = q = 1.

Now, using part (2) of Theorem 1 and the fact that K†
+ = K−, we get the next theorem.

Theorem 2. The defining relations of lp,q can be either:

[K0,K+]p,q = rK+, and [K+,K−]p,q = F (K0) (7)

or in its Hermitian form,

[K−,K0]p,q = rK−, and [K+,K−]p,q = F (K0) .

Proof. Consider the Hermitian conjugate of (4). From part (2) of Theorem 1 and the

facts that K†
+ = K− and K†

0 = K0, we have
(
[K0,K+]p,q

)†
=

[
K†

+,K
†
0

]
p,q

=
[
K†

−,K0

]
p,q

=

(rK+)
† = rK−. So, we have the second defining relation of lp,q, (5). Alternatively, (4) is

actually the Hermitian conjugate of (5).



L. A-M. Hanna, S. S. Hassan, M. Almutairi / Eur. J. Pure Appl. Math, 18 (4) (2025), 6733 5 of 12

4. Faithful representations of lp,q

Some basic and necessary definitions are presented at the beginning of this section.

Definition 2. Let K be a field and Mn (K) be the set of all n × n matrices of entries
from K. A matrix representation of the degree n of the (p, q)-deformed Lie algebra, lp,q,
is a mapping ρ : lp,q → Mn (K) satisfying the following properties, for all u and v in lp,q
and all α and β in K:

(i) ρ (u+ v) = ρ (u) + ρ (v) ,

(ii) ρ (uv) = ρ (u) ρ (v) ,

(iii) ρ (αu) = αρ (v), and

(iv) ρ
(
[u,v]p,q

)
= pρ (uv)− qρ (vu) .

The matrix ρ (u) is called the representation matrix of u in lp,q. If ρ is a one-to-one
mapping, then the representation is said to be faithful.

It can be shown that the representation matrices of a linearly independent set of ele-
ments in lp,q are linearly independent. Here, we assume that A,B, and C are representation
matrices for the generators of lp,q, namely, K+,K−, and K0, respectively. All representa-
tions under consideration are supposed to satisfy the physical properties: B = A†, and C
is a real diagonal matrix. Also, p, q, r ∈ R∗ and F (x) is a polynomial function in R [x].
Faithful matrix representations of lp,q of the least degree are the main purpose of this
work. Mind that the representation matrices A,B, and C are supposed to be linearly
independent in the case of faithful representations. Also, we use 0 for the zero matrix of
appropriate size, and o for the zero element of lp,q.

For a representation of degree n, the following equations (8)-(12) are necessary relations
for A,B, and C, which are obtained from (4) and (6), respectively. For i, j = 1, 2, ..., n,
we have

[r − (pcii − qcjj)] aij = 0 for i 6= j, (8)

aii [r − (p− q) cii] = 0, (9)

and ∑n

t=1
(paitājt − qatj āti) = 0 for i 6= j, (10)∑n

t=1

(
p |ait|2 − q |ati|2

)
= F (cii) , (11)

(p− q)
∑n

i=1

∑n

t=1
|ait|2 =

∑n

i=1
F (cii) . (12)



L. A-M. Hanna, S. S. Hassan, M. Almutairi / Eur. J. Pure Appl. Math, 18 (4) (2025), 6733 6 of 12

5. Faithful matrix representations of lp,q

Since lp,q is generated by 3 generators, namely, K± and K0, then the least possible
degree of a faithful matrix representation is 2. So, let

A =

[
a b
c d

]
, B = A† and C = diag (c1, c2) , where c1, c2 ∈ R, while a, b, c, d ∈ C (13)

be linearly independent 2× 2 matrices.
Thus, from (13) in (6), we have,
[A,B]p,q = pAA† − qA†A

= p

[
a b
c d

] [
ā c̄
b̄ d̄

]
− q

[
ā c̄
b̄ d̄

] [
a b
c d

]
= p

[
|a|2 + |b|2 ac̄+ bd̄

āc+ b̄d |c|2 + |d|2
]
− q

[
|a|2 + |c|2 āb+ c̄d

ab̄+ cd̄ |b|2 + |d|2
]

=

 (p− q) |a|2 + p |b|2 − q |c|2 p
(
ac̄+ bd̄

)
− q (āb+ c̄d)

p
(
ac̄+ bd̄

)
− q (āb+ c̄d) p |c|2 − q |b|2 + (p− q) |d|2


= F (C) .
Thus, we have

[A,B]p,q =

 (p− q) |a|2 + p |b|2 − q |c|2 p
(
ac̄+ bd̄

)
− q (āb+ c̄d)

p
(
ac̄+ bd̄

)
− q (āb+ c̄d) p |c|2 − q |b|2 + (p− q) |d|2


= diag (F (c1) , F (c2)) . (14)

Similarly, using (13) in (4), we have,

[C,A]p,q = p

[
c1 0
0 c2

] [
a b
c d

]
− q

[
a b
c d

] [
c1 0
0 c2

]
= rA.

So, we have,

[C,A]p,q =

 (p− q) c1a (pc1 − qc2) b

(pc2 − qc1) c (p− q) c2d

 = rA. (15)

From (14), as tr
(
[A,B]p,q

)
= F (c1) + F (c2) .

Thus, we have,

(p− q)
(
|a|2 + |b|2 + |c|2 + |d|2

)
= F (c1) + F (c2) . (16)

Similarly, from (15), we have, tr
(
[C,A]p,q

)
= ra+ rd.

So,
(p− q) (ac1 + dc2) = r (a+ d) . (17)



L. A-M. Hanna, S. S. Hassan, M. Almutairi / Eur. J. Pure Appl. Math, 18 (4) (2025), 6733 7 of 12

Lemma 1. If C = kI2, a scalar matrix and p 6= q, then k = r
p−q .

Proof. Since [C,A]p,q = (p− q) kA = rA. Since A 6= 0, then k = r
p−q .

5.1. Representation matrices of degree 2 of lq,q

In this subsection, we consider the special case when p = q.

Theorem 3. lp,q is a Lie algebra iff p = q.

Proof. Let x,y ∈ lq,q, then from definition (1), we have [x,y]q,q = − [y,x]q,q and
[x,x]q,q = o. From parts (4) and (5), the bilinearity is satisfied, while the Jacobi identity
is satisfied from part (9) of Theorem 1, respectively.

Conversely, from definition (1), ∀x ∈ lp,q, we have [x,x]p,q = (p− q)x2 = o, only if
p = q or x2 = o, ∀x ∈lp,q. Since C is a diagonal matrix, then C2 = 0, if and only if,
C = 0, which is impossible since C is a representation matrix of a basis element of lp,q.

As lq.q is a Lie algebra, we get the following Corollary.

Corollary 1. In lq,q, tr (A) = tr (F (C)) = 0.

Proof. From part (1) of Theorem 1, as p = q, we have tr ([X,Y ]) = 0 for every
X,Y ∈ lq,q, then the corollary results from (14) and (15).

Lemma 2. If C is a scalar matrix, then the matrix representation of lq,q is not faithful.

Proof. Let C be a scalar matrix, then from (15), as p = q and since r 6= 0, then A = 0.

Lemma 3. For faithful representation of lq,q, the representation matrix A =

[
0 b
0 0

]
,

where b is a nonzero complex number.

Proof. From (9), if i = 1, we have a [r − (q − q) c1] = 0, then a = 0 and similarly, if
i = 2, we have d = 0. From (8), if i = 1 and j = 2, we have b [r − q (c1 − c2)] = 0, and
similarly, for i = 2 and j = 1, we have c [r − q (c2 − c1)] = 0. Thus, suppose bc 6= 0, then
we have q (c1 − c2) = r = q (c2 − c1). As q 6= 0, one gets that c1 = c2, i.e., C is a scalar
matrix. From Lemma 2, the representation is not faithful. Therefore, bc = 0. The case
where b = c = 0, implies that A = 0 which is rejected since A is a basis element. Hence,
the lemma.

Theorem 4. The Lie algebra lq,q, where q ∈ R∗, has a faithful representation of degree 2 as
the least degree, if and only if, there exists t ∈ R, such that F (t) = −F

(
t− r

q

)
with F (t)

q >

0. Moreover, the representation matrices of K+,K−, and K0 are A =

[
0 b
0 0

]
, B = A†,

and C = diag
(
t, t− r

q

)
, respectively, such that |b|2 = F (t)

q .



L. A-M. Hanna, S. S. Hassan, M. Almutairi / Eur. J. Pure Appl. Math, 18 (4) (2025), 6733 8 of 12

Proof. From Lemma 3, A =

[
0 b
0 0

]
, with b 6= 0, that is |b|2 > 0. From (15), we have

bq (c1 − c2) = rb. Thus, c2 = c1 − r
q , since b 6= 0.

From (14), F (c1) = q |b|2, and F (c2) = −q |b|2 = −F (c1). Thus, for a faithful
representation of lq,q, the polynomial function F should satisfy that F (t) = −F

(
t− r

q

)
for some real number t. Actually, c1 = t. Thus C = diag

(
t, t− r

q

)
. If there is no

such t, then the representation matrix C cannot be found and hence, lq,q has no matrix
representation. Also, from (14), we have |b|2 = F (t)

q must be positive, because if it is
negative, A does not exist, and if it is 0, then A = 0, and the representation is not
faithful. Hence the theorem.

The following examples demonstrate the method for calculating the representation
matrices.

Example 1. Given that p = q = 2, r = 4 and F (x) = x3 + x. First we consider
the equation F

(
t− r

q

)
= −F (t). Thus, (t− 2)3 + (t− 2) = −

(
t3 + t

)
, which is 2t3 −

5t2 + 13t − 10 = 0. Choose the real solution t = 1, as it satisfies that F (t)
q > 0, since

F (t)
q = 13+1

2 = 1 > 0. So, |b|2 = F (t)
q = 1. Take b = 1. Therefore, A =

[
0 1
0 0

]
, B = A†

and C = diag (1,−1) are representation matrices of K+,K− and K0, respectively. Since

[A,B]2,2 = 2

[
0 1
0 0

] [
0 0
1 0

]
−2

[
0 0
1 0

] [
0 1
0 0

]
=

[
2 0
0 −2

]
= diag (F (1) , F (−1)) .

And
[C,A]2,2 = 2

[
1 0
0 −1

] [
0 1
0 0

]
− 2

[
0 1
0 0

] [
1 0
0 −1

]
=

[
0 4
0 0

]
= 4A.

Clearly, the representation is faithful, because the matrices A,B, and C are linearly
independent.

Another representation matrices can be considered as |b|2 = 1, let A =

[
0 i
0 0

]
, B =

A† and C = diag (1,−1). Such a representation does not satisfy the condition K+ +K−
is real. Actually, this condition is only satisfied when A is a real matrix.

Example 2. Let F (x) = x2+x+1. Then lq,q has no faithful representation since F (t) > 0
for any t ∈ R.

Example 3. The model of light amplifier, namely, [K+,K−] = −2K0, [K0,K±] = ±K±

[17], where p = q = 1, r = 1 and F (x) = −2x. Solving F (t) = −F
(
t− r

q

)
for t, we

get t = 1
2 is the only real solution, but F (t)

q =
−2

(
1
2

)
1 = −1 not positive, i.e., |b|2 < 0,

which is impossible. So, A does not exist. So, the light amplifier model has no matrix
representation satisfying the physical conditions, as previously proved in [17].



L. A-M. Hanna, S. S. Hassan, M. Almutairi / Eur. J. Pure Appl. Math, 18 (4) (2025), 6733 9 of 12

5.2. Representation matrices of degree 2 of l−q,q

In this subsection, we consider the special case when p = −q.

Lemma 4. In l−q,q, if C is a scalar matrix, then A is of zero diagonal elements.

Proof. Let A as in (13), and C is a scalar matrix. So, let C = tI2. For the linear inde-
pendence of the generators, and since t is a scalar matrix, we assume that d = 0. Suppose
a 6= 0. We have from (14), F (t) = −q

(
2 |a|2 + |b|2 + |c|2

)
and F (t) = −q

(
|b|2 + |c|2

)
.

Then a = 0 as q 6= 0.

Theorem 5. The l−q,q has faithful matrix representation with C a scalar matrix, if F (x)

satisfies, that −1
qF

(
r

−2q

)
> 0, and the representation matrices of K+,K−, and K0 are,

A,A†, and C, respectively, where

(i) A =

[
0 b
c 0

]
, A†, and C = r

−2q I2, respectively, where |b|2 + |c|2 = −1
qF

(
r

−2q

)
> 0

such that |b|2 6= |c|2, with b and c are of equal imaginary parts to satisfy that K++K−
is real.

(ii) A =

[
0 b
0 0

]
, A†, and C = r

−2q I2, respectively, where |b|2 = −1
qF

(
r

−2q

)
> 0. In

this representation, the condition K+ + K− is real, and will only be satisfied for a
real matrix A.

Proof. Let C = tI2 scalar matrix and from Lemma 4, a = d = 0. Thus A =

[
0 b
c 0

]
.

From (14), −q
(
|b|2 + |c|2

)
= F (t).

Consider the two cases. Case 1: If bc = 0. Let b 6= 0, c = 0. Then from (8),
(r + 2qt) = 0. Hence, t = r

−2q .

Then F
(

r
−2q

)
= −q |b|2. Therefore, |b|2 = −1

qF
(

r
−2q

)
> 0.

Case 2: If bc 6= 0, then from (8), we have b (r + 2qt) = 0 and c (r + 2qt) = 0. Then
t = r

−2q . Then |b|2 + |c|2 = −1
qF

(
r

−2q

)
> 0.

To satisfy that K+ +K− is a real operator, one must have, b+ c̄ ∈ R.
If |b|2 = |c|2, then A and B are linearly dependent, and hence the representation is not

faithful.

Example 4. For l−2,2, if r = 8 and F (x) = x−18, then C = −2I2. To get a representation
of type in 1, we can take, for instance, b = 3 and c = ±1, since |b|2 + |c|2 = 10.

Also, we can take b =
√
2 + 2i and c = 2i.

Example 5. For l−2,2, if r = 8 and F (x) = x− 18, then C = −2I2 and take b = ±
√
10,

for a representation of type in 2.



L. A-M. Hanna, S. S. Hassan, M. Almutairi / Eur. J. Pure Appl. Math, 18 (4) (2025), 6733 10 of 12

Now consider the case where C = diag (c1, c2), which is not a scalar matrix. From (14)
and (15), we have the following theorem as p = −q.

Theorem 6. Let p = −q. Then the l−q,q has faithful matrix representations, with C not
a scalar matrix, namely,

(i) if there exists t ∈ R, such that F (t) = −q
(
|b|2 + |c|2

)
and F

(
−qt−r

q

)
= F (t) where

C = diag
(
t,− qt+r

q

)
and A =

[
0 b
c 0

]
, where |c|2 = − q|b|2+F (t)

q > 0,

and for the special case, when c = 0,

(ii) if there exists t ∈ R, such that |b|2 = −F (t)
q > 0 and F

(
− qt+r

q

)
= F (t) where

C = diag
(
t,− qt+r

q

)
and A =

[
0 b
0 0

]
.

Proof. Let p = −q, then we have from (14), with a = d = 0, we get F (c1) = F (c2) =

−q
(
|b|2 + |c|2

)
. So, Let c1 = t, then |c|2 = − q|b|2+F (t)

q > 0, if bc 6= 0. From (15), we have,
(−qt− qc2) b = rb, thus, if bc 6= 0, then c2 = − qt+r

q .

5.3. Representation matrices of degree 2 of lp,q where p2 6= q2

In this subsection, we consider the case where p2 6= q2, and we seek a faithful repre-
sentation for lp,q of degree 2 as the least degree.

Lemma 5. If bc 6= 0, and p+ q 6= 0, then C is a scalar matrix.

Proof. Comparing the elements on both sides of equation (15), we get:
if b 6= 0, then, r = pc1 − qc2 and similarly, when c 6= 0, then r = pc2 − qc1. Thus,

(p+ q) (c1 − c2) = 0. Therefore, C is a scalar matrix.

Theorem 7. If C is a scalar matrix and p+ q 6= 0, then the representation of lp,q is not
faithful.

Proof. For the linear independence of the generators, and as C is a scalar matrix, we
assume that d = 0. So, from (14), we have p (ac̄) − q (āb) = 0. Suppose a 6= 0, then
b =

pc̄a

qā
and hence |b|2 = p2|c|2

q2
. Since C is scalar, then F (c1) = F (c2) , and hence from

(14), (p− q) |a|2 + p |b|2 − q |c|2 = p |c|2 − q |b|2. If p + q 6= 0, we have, |c|2 = − q2|a|2

(p+q)2

which is impossible unless c = 0, and in such a case a = 0, contradicting with a 6= 0. So,
a = 0. Then from (14), we get F (c1) = p |b|2 − q |c|2 and similarly, F (c2) = p |c|2 − q |b|2.
Therefore, as C is scalar, we get (p+ q)

(
|b|2 − |c|2

)
= 0. Then either A = 0 or |b|2 = |c|2.

In both cases, the representation is not faithful, since the generators A,B = A†, and C
are linearly independent, and A+B is a real matrix.



L. A-M. Hanna, S. S. Hassan, M. Almutairi / Eur. J. Pure Appl. Math, 18 (4) (2025), 6733 11 of 12

Theorem 8. The Lie algebra lp,q, where p2 6= q2 with pq 6= 0, has a faithful representation
of degree 2 as the least degree, iff there exists t ∈ R, such that F (t)

p > 0 and F
(
pt−r
q

)
=

− q
pF (t).

Moreover, the representation matrices of the generators K+,K−, and K0 are A,A†,

and C = diag
(
t, pt−r

q

)
, respectively, where A =

[
0 b
0 0

]
, such that |b|2 = F (t)

p > 0.

Proof. From Lemma 5 and Theorem 7, we must have that bc = 0. So, we can choose

A =

[
0 b
0 0

]
, with b 6= 0. Thus, from (14), we have F (c1) = p |b|2 and F (c2) = −q |b|2.

Thus, |b|2 = F (c1)
p = −F (c2)

q . From (15), we have (pc1 − qc2) = r. So, pc1− r = qc2. Thus,
c2 =

pc1−r
q . Let c1 = t a real number, then c2 =

pt−r
q .

Thus, |b|2 = F (t)
p = −

F
(

pt−r
q

)
q > 0. Therefore, the function F should satisfy that

F
(
pt−r
q

)
= − q

pF (t) .

Example 6. For l2,3, if r = −1 and F (x) = 3x2 − x, then the solutions of the equation
F
(
pt−r
q

)
= − q

pF (t), are t = 0 or t = 1
7 .

For t = 0, we have |b|2 = 0, rejected.
For t = 1

7 , we have |b|2 = − 2
49 < 0, rejected.

Example 7. For l2,3, if r = −1 and F (x) = 2x2 − x, then the solutions of the equation
F
(
pt−r
q

)
= − q

pF (t), are t = 2
5 or t = − 1

14 .

For t = − 1
14 , |b|

2 = 2
49 > 0 and C = diag

(
− 1

14 ,
2
7

)
.

For t = 2
5 , |b|

2 = − 1
25 < 0, rejected.

6. Conclusion

Our main purpose in this work was to find faithful matrix representations of the two-
parameter (p, q)-deformed Lie algebra lp,q defined in (4)-(6) with p and q being nonzero
real numbers. The derived representations associated with the operator generators K±,0

in the Hamiltonian model (3) can be utilized to investigate the quantum state evolution,
similar to our earlier work in the case of ordinary Lie bracket, where p = q = 1 [10].

Acknowledgements

The authors are grateful to Kuwait University.

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