# On Faithful Matrix Representations ofq-Deformed Models in Quantum Optics

Authors: Latif A -M. Hanna; Abdullah Alazemi; Anwar Al-Dhafeeri
Journal: International Journal of Mathematics and Mathematical Sciences
 (2022)

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## Abstract

Consider theq-deformed Lie algebra, tq:K^1,K^2q=1−qK^1K^2,K^3,K^1q=sK^3, K^1,K^4q=sK^4,K^3,K^2q=tK^3,K^2,K^4q=tK^4, and K^4,K^3q=rK^1, where r,s,t∈ℝ−0, subject to the physical properties: K^1 and K^2 are real diagonal operators, and K^3=K^4†, († is for Hermitian conjugation). The q-deformed Lie algebra, tq is introduced as a generalized model of the Tavis–Cummings model (Tavis and Cummings 1968, Bashir and Sebawe Abdalla 1995), namely, K^1,K^2=0,K^1,K^3=−2K^3,K^1,K^4=2K^4,K^2,K^3=K^3,K^2,K^4=K^4, and K^4,K^3=K^1, which is subject to the physical properties K^1 and K^2 are real diagonal operators, and K^3=K^4†. Faithful matrix representations of the least degree of tq are discussed, and conditions are given to guarantee the existence of the faithful representations.


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## Body

## 1. Introduction

In the field of quantum optics, the Tavis–Cummings (TC) model [1] involves the multiphoton interaction of N>1 2-level atoms with a single mode quantized field, in the absence of any radiation damping. It is a generalization of the Jaynes–Cummings (JC) model [2] of a single 2-level atom interacting via single photon transition with a single quantized field mode. The Lie algebra approach in [3–5] was employed in [6], where the Hamiltonian description of the the TC model of a coupled system of a boson (single mode of the radiation field) and N fermions (N two-level atoms) is given by H=ω0c^†c^+1/2ω1J^z+gc^†mJ^−+c^mJ^+, where c^ and c^† are the boson field annihilation and creation operators satisfying c^,c^†=1, while J^± are the usual collective angular momentum operators, with ω0 and ω1 denoting the field and atomic frequencies, while g and m are the coupling constant and the number of photons, respectively. Using the transformation in [7–9], c^m=fn^cb^, and c^†m=b^†fn^c, the operators b^ and b^† are boson operators satisfying the commutation relation b^i,b^j†=δij, and the functions fn^c are real functions of the number operator n^c=c^c^† given by fn^c=n^c+m!/n^c!n^c/m+1!1/2 , and on using the Holstien–Primakoff transformation [10], J^−=J^−n^aa^eiϕt,J^+=a^†J^−n^ae−iϕt, and n^a=a^†a^=1/2J^+J^z, where the phase ϕt is an arbitrary function of time, the Hamiltonian becomes(1)H=ω1a^†a^+ω2b^†b^+gb^†fn^ca^J^−n^aeiϕt+c.c.,where ω2=mω0. On approximating n^c and n^a by their C-number time-dependent functions nct and nat, then the Hamiltonian (1) is bilinearized and has the form(2)H=ω1a^†a^+ω2b^†b^+λta^b^†eiϕt+a^†b^e−iϕt,where λt=gJ^−n^atfn^ct, J^≥n^a. In this case, the Hamiltonian (2) can be regarded as that of two-mode coupled oscillators with a time-dependent pumping and phase [11–15].As in [4, 5], define operators K^l,K^2,K^3,K^4 as follows: K^1=a^†a^−b^†b^,K^2=b^†b^,K^3=a^b^†,K^4=a^†b^, which satisfy the following commutation relations:(3)K^1,K^2=0,K^1,K^3=−2K^3,K^1,K^4=2K^4,K^2,K^3=K^3,K^2,K^4=K^4,K^4,K^3=K^1.Then, the Hamiltonian (2) becomes(4)H=ω1K^1+ω1+ω2K^2+λtK^3eiϕ+K^4e−iϕ.In [6], a matrix representation for the operators K^1,K^2,K^3 and K^4 satisfying (3) is found to be K^1=100−1,K^2=0001,K^3=0010,K^4=0100.In [16], two optical atom models were introduced, namely, the model of the two-level optical atom as follows:(5)K^+,K^−=K^0,K^0,K^±=±2K^±,and the model of a light amplifier, which corresponds to the su1,1 algebra(6)K^+,K^−=−2K^0,K^0,K^±=±K^±.The Hamiltonian of these two coupled quantized harmonic oscillators models is of the following form:(7)H=ωa^†a^−b^†b^+λTa^†b^+a^b^†,where λT is an arbitrary time-dependent coupling parameter, and the Boson operators of the two-level optical atom obey the commutation relations:(8)a^,a^†=1=b^,b^†.As a combination of the three operators, namely,K^0,K^±, the Hamiltonian can be written as(9)H=ωK^0+λTK^++K^−,where K^0=a^†a^−b^†b^, and K^+=a^†b^=K^−†, with K^++K^− must be real to satisfy the Hermiticity of the Hamiltonian H. In [17], it was shown that the Lie algebra generated by the three operators, K^0,K^± is the su2 algebra. In [18, 19], faithful matrix representations of the least degree for the two-level optical atom were found to be(10)K^+=0100,K^−=K^+†=0010,K^0=100−1,while the model of a light amplifier has no faithful representation satisfying the physical requirements.In fact, faithful matrix representations of optical models give more understanding of such models. They are also useful to visualize its properties in algebraic ways. The least degree faithful representation is an ideal tool to handle such models. Steinberg, in [3, 20], introduced a method to solve partial differential equations. The method utilizes faithful matrix representation of the least degree. The method was applied in [21] to solve the Schrödinger’s wave equation of the q-deformed coupled quantized oscillator. The q-deformed Lie algebras are introduced for more essential generalizations in quantum mechanics. For instance, [22], where in [23–26], it was considered for q to be in 0,1, for the particular model of Fermion oscillators.Definition 1.
LetX and Y be n×n matrices. The q-deformed Lie bracket of X and Y is defined to be(11)X,Yq=XY−qYX,forq∈ℝ.
On generalizing such optical models, the given representation matrices cover more models. Particularly, when they share the same mathematical properties. For instance, the TC model (3), K^1,K^4, and K^3 satisfy the same commutation relations of K^0,K^+, and K^−, respectively, of the models in (5) and (6), apart from the scalar coefficients, which will be considered by r,s, and t, in the general case. Also, the q-deformed Lie algebras offer more generalization.
In this paper, theq-deformed Lie algebra, tq is introduced as a generalization of the Tavis–Cummings model in (3), namely,(12)tq:K^1,K^2q=1−qK^1K^2,K^3,K^1q=sK^3,K^1,K^4q=sK^4,K^3,K^2q=tK^3,K^2,K^4q=tK^4,K^4,K^3q=rK^1,subject to the physical requirements, K^3=K^4†, and K^1 and K^2 are real diagonal operators.
The main purpose of this paper is to construct matrices of the least degree faithfully representing theq-deformed Lie algebra in (12). The q-deformed Lie algebra in (12) is chosen as a generalization of the TC model in (3). In abstract point of view, (3) can be considered as an extension of the models of (5) and (6), by the operator K^2. Throughout this paper, A,A†,B, and C are the representation matrices of K^4,K^3,K^1, and K^2, respectively, of the model (12), with B and C as real diagonal matrices. On the other hand, A,A†, and B can represent K^+,K^−, and K^0, respectively, in models (5) and (6), where B is the real diagonal matrix.
So, the main objective of this paper is to provide, if any, linearly independent matricesA,A†,B, and C, satisfying(13)tq:B,Cq=1−qBC,A†,Bq=sA†,B,Aq=sA,A†,Cq=tA†,C,Aq=tA,A,A†q=rB.
Faithful matrix representations of degree 2 were proved to be the least degree. Also,tq was given as an extension of the generalized model of the q-deformed couple quantized harmonic oscillator by the operator K^2.
The representation matrices, fortq, of degree 2 as the least degree, which satisfy the physical conditions, are given in Theorem 5 as follows.
Forq∉−1,0, and a,b∈ℝ, such that a2≤rs/1+q2, we have(14)A=0a±irs−1+q2a21+q200,A†=00a∓irs−1+q2a21+q20,B=s1+q2100−q,C=t+bq00b.
The special cases whereq=0 and q=−1 are given in Theorems 5 and 6, respectively.


## 2. Preliminaries

LetX and Y be n×n matrices. The q-deformed Lie bracket of X and Y is defined, [11] to be X,Yq=XY−qYX, for q∈ℝ. It should be noticed that for q=0, the q-deformed Lie bracket is the ordinary matrix multiplication of X and Y, whereas for q=1, it is the ordinary Lie bracket. Thus, we always write X,Y1 as X,Y, whenever needed. The matrix representation of the TC model were given for q=1 in [24]. So, the cases where q∈ℝ−0,1 will always be in our interest.Unless otherwise stated, the square matricesA,A†,B, and C denote the representation matrices of the operators K^4,K^3,K^1, and K^2, respectively, of the model (12), and 0 is the zero square matrix of an appropriate size. The set ℕn denotes the set 1,2,…n for positive integer n. All representations are supposed to satisfy the physical requirements that K^1 and K^2 are real diagonal operators. Also, r,s,t∈ℝ such that rst≠0.


## 3. Basic Properties of theq-Deformed Lie Bracket

Fori,j∈ℕn, let eij be the matrix unit, which is an n×n matrix whose element in the ith row and jth column is 1 and zero elsewhere. In other words, the eij,s are the elements of the standard basis of Mn×nℂ, the set of all n×n matrices over the field ℂ of complex numbers. Since(15)eijekl=δjkeil,where δij is the Kronecker delta, we have the following lemma.Lemma 1.
Consider the matrix units,eij for i,j∈ℕn, then for q∈ℝ, we have(16)eij,eklq=δjkeil−qδilekj.(1)
Ifj≠k and l≠i, then eij,eklq=0.(2)
Ifi≠j, then eij,eijq=0.(3)
Ifl≠i, then eij,ejlq=eil.(4)
Ifj≠k, then eij,ekiq=−qekj.(5)
Ifi≠j, then eij,ejiq=eii−qejj.(6)
eii,eiiq=1−qeii.Proof.
From (11) and (15), we have eij,eklq=eijekl−qekleij=δjkeil−qδilekj. Other parts are easily driven from (16).
To be self-contained, we copy the following properties of theq-deformed Lie bracket from [25] in Theorem 1. We also, add other properties in Lemma 2.Theorem 1.
LetMn×nℂ be the set of all n×n matrices over a field ℂ and q∈ℝ. Let α,β∈ℂ. Then, for X,Y and Z in Mn×nℂ, we have(1)
trX,Yq=trY,Xq.(2)
X,Yq†=Y†,X†q.(3)
X+Y,Zq=X,Zq+Y,Zq.(4)
X,Y+Zq=X,Yq+X,Zq.(5)
X,Yq=qX,Y+1−qXY.(6)
αA,βBq=αβA,Bq.(7)
X,Yq=X,Y+1−qYX.(8)
X,Xq=1−qX2.(9)
Y,Xq=X,Yq−1+qX,Y.(10)
X,Y,Zqq+Y,Z,Xqq+Z,X,Yqq=1−qXYZ+YZX+ZXY−qXZY+YXZ+ZYX.Part 10 is theq-deformed version of the Jacobi identity. So, we add here the following properties.Lemma 2.
LetMn×nℂ be the set of all n×n matrices over ℂ and q∈ℝ. Let α,β∈ℂ. Then, for X,Y, and Z in Mn×nℂ, we have(1)
XY,Zq=XY,Z+1−qZXY=1−qXYZ+qXY,Z and also,(2)
XY,Zq=X,YZ+Y,ZXq.(3)
X,YZq=X,YZ+1−qYZX=1−qXYZ+qX,YZ and also,(4)
X,YZq=XY,Zq+qZX,Y.(5)
X,Yq,Zq+Y,Zq,Xq+Z,Xq,Yq=1−qXYZ+YZX+ZXY−qXZY+YXZ+ZYX.Proof.
For part (1), we have XY,Zq=XYZ−qZXY=XYZ+ZXY−ZXY−qZXY=XY,Z+1−qZXY. Also, XY,Zq=XYZ−qZXY=XYZ−qXYZ+qXYZ−qZXY=1−qXYZ+qXY,Z. For part (2), XY,Zq=XYZ−qZXY=XYZ−qYZX+qYZX−qZXY=X,YZq+qY,ZX.
X,YZq=XYZ−qYZX=XYZ−YZX+YZX−qYZX=X,YZ+1−qYZX. Similarly, for parts (3) and (4), Part (5) is the q-deformed version of the Jacobi identity that can be easily driven from the definition (11).


## 4. Properties of the Representation Matrices

Suppose thattq has a faithful matrix representation of degree n, where the n×n matrices A,A†,B, and C are representation matrices of K^4,K^3,K^1, and K^2, respectively, of the model (12). Thus, A,A†,B, and C are linearly independent with B and C as real diagonal matrices.Lemma 3.
Letu,v∈ℕn, and α=u,v be a transposition. The representation obtained by applying α to the rows as well as to the columns of matrices A,A†,B, and C is a conjugate representation for tq and satisfies the physical requirements.Proof.
LetX be the elementary matrix obtained by applying α to the rows of In. Since X=X−1=XT=X†=X−1†, then A′=X−1AX,A′†=X−1AX†=X†A†X−1†=X−1A†X,B′=X−1BX, and C′=X−1CX are representation matrices for K^4,K^3,K^1, and K^2, respectively, of the model (12), satisfying A′,A′†q=X−1AX,X−1A†Xq=X−1AXX−1A†X−qX−1A†XX−1AX=X−1AA†−qA†AX=X−1A,A†qX=X−1rBX=rB′. Similarly, B′,A′q=X−1B,AqX=X−1sAX=sA′ and C′,A′q=tA′. Also, B′,C′q=X−1B,CqX.
Astq is defined to be a generalization of the Tavis–Cummings model in (3), it should be pointed out that the q-deformed commutator K^1,K^2q cannot be equal to 0 for q≠1.Theorem 2.
The matrix productBC≠0. Moreover, the commutator B,Cq=0 if q=1.Proof.
LetA,B, and C be representation matrices of K^4,K^1, and K^2, respectively, in (3). Suppose that B,Cq=0, and q≠1. Since B and C are diagonal, then B,Cq=1−qBC=0 if BC=0. Thus, if Bii≠0, then Cii=0 for i∈ℕn. Using Lemma 3, to rearrange B such that, B=diagT,V, where T is an l×l nonsingular diagonal matrix, and V=0n−l×n−l matrix, while C=diag0l×l,Sn−l×n−l, for some l∈ℕn.
Since the diagonal elements ofB are all zeros for each i=l+1,l+2,…,n; we have sAii=B,Aqii=BAii−qABii=0−q0=0. So, Aii=0 for i>l. Similarly, since for each i=1,…,l; we have C,Aii=CA−qACii=0=tAii. Hence, Aii=0 for i≤l. Thus, all diagonal elements of A are zeros.
SinceA≠0, otherwise the representation is not faithful, then for some i≠j, there is an Aij≠0. If i>l, then the ith row and the ith column of B are zero row and zero column, respectively. As B,Aq=sA, thenB,Aqij=∑t=1nBitAtj−qAitBtj=0, contradicting that sAij≠0, as s≠0. Similarly, if i≤l, then the ith row and the ith column of C are zero row and zero column, respectively, and as C,Aq=tA, then C,Aqij=0 contradicting that tAij≠0, as t≠0.
AsB and C are diagonal matrices, then their commutator B,Cq=1−qBC is too trivial to be included in the defining relations of tq. Then, we have the following theorem.Theorem 3.
The defining relations oftq can be either(17)tq:A†,Bq=sA†,B,Aq=sA,A†,Cq=tA†,C,Aq=tA,A,A†q=rB,or(18)tq:A,A†q=rB,B,Aq=sA,C,Aq=tA.Subject to the physical conditions:B and C are real diagonal operators.Proof.
From part 2 of Theorem1, we have B,Aq†=sA†, which implies that A†,Bq=sA†. Similarly, taking the Hermitian conjugate of both sides of the commutator, C,Aq=tA, we get A†,Cq=tA†.


## 5. Least Degree Faithful Matrix Representation of theq-Deformed Tavis–Cummings Model

We first try to investigate whethertq has faithful representations of degree 2. In this section, unless otherwise stated, let A=ae11+be12+ce21+de22, A†,B=fe11+ge22, and C=ue11+ve22 be representation matrices of K^4,K^3,K^1, and K^2 respectively in (12), where a,b,c,d∈ℂ and f,g,u,v∈ℝ, satisfying that for r,s,t∈ℝ−0,(19)A,A†q=rB,(20)B,Aq=sA,(21)C,Aq=tA,where the matrix units eij’s are 2×2 matrices for i,j∈ℕ2.Theorem 4.
Consider the2×2 matrix units, e11,e12,e21, and e22. Let A=ae11+be12+ce21+de22, A†,B=fe11+ge22, and C=ue11+ve22 be representation matrices of K^4,K^3,K^1, and K^2, respectively, where a,b,c,d∈ℂ and f,g,u,v∈ℝ. Then, for r,s,t∈ℝ−0, we have(1)
ac¯+bd¯−qba¯+dc¯=0,(2)
1−qa2+b2−qc2=rf,(3)
1−qd2+c2−qb2=rg,(4)
trB=f+g=1−q/ra2+b2+c2+d2,(5)
1−qfa=sa,(6)
1−qgd=sd,(7)
f−qgb=sb,(8)
g−qfc=sc,(9)
1−qua=ta,(10)
1−qvd=td,(11)
u−qvb=tb,(12)
v−quc=tc.Proof.
Using parts (3) and (4) of Theorem 1 and using Lemma 1, in (19), and from the linear independence of the matrix units, we get parts (1)–(3). Similarly, from (20), we get parts (5)–(8). From (21), we get parts (9) and (12) in the same way. From parts (2) and (3) of the theorem, we get part (4).
Now, we use Theorem4, in order to build the representation matrices for tq and give conditions to guarantee the faithful representations. But first we consider the special cases when q=0 and q=−1 in Theorems 5 and 6, respectively.Theorem 5.
Ifq=0, then the 0-deformed Lie algebra t0 has faithful representations of degree 2, as the least degree. The representation matrices of the generators K^4,K^3,K^1, and K^2 are, respectively, for x,y,m∈ℝ, such that 0≤y2≤rs−x2 and rstm≠0,(1)
A=x+iy±rs−x2+y200,A†=x−iy0±rs−x2+y20,B=s000, and C=t00m, or(2)
A=±rs−x2+y2x+iy00,A†=±rs−x2+y20x−iy0,B=s000, and C=t00m.Proof.
Letq=0. If ab≠0, then from part (5) in Theorem 4, we have f=s, then we get from part (2) that b2=rs−a2. Thus, if a=x+iy, then b2=rs−x2+y2. Thus, b=±rs−x2+y2. For faithful representation, take m≠0. Cases, if ac≠0,ad≠0,bc≠0, or bd≠0 lead to representations where K^1 and K^2 are represented by scalar matrices, and hence, the representations are not faithful. If cd≠0, then the representation is conjugate to either of the given representations.
A real representation can be obtained by takingy=0 in Theorem 5. A practically useful representation is when A is a nilpotent matrix, i.e., the case when x=y=0. So, we have the following corollary.Corollary 1.
Ifq=0, then the 0-deformed Lie algebra t0 is isomorphic to gl2,ℝ and has faithful representations of degree 2 as the least degree if and only if rs>0. The representation matrices of the generators K^4,K^3,K^1, and K^2 in (12), are, respectively, for any m∈ℝ−0,(22)A=0±rs00,A†=00±rs0,B=s000,C=t00m.Theorem 6.
Ifq=−1, then the −1-deformed Lie algebra t−1 has faithful representations of degree 2 as the least degree. The representation matrices of the generators K^4,K^3,K^1, and K^2 in (12), are, respectively,(23)A=0x+iy±rs2−x2+y20,A†=0±rs2−x2+y2x−iy0,B=s2I2,C=m00t−m,where x,y∈ℝ, and m≠t/2 such that x2+y2≤rs/2.Proof.
Letq=−1. We claim that we should have a=0. The case where a≠0 and b=c=0 leads to the representation matrix A is diagonal and then, the representation is not faithful.
Ifab≠0, then from part (5) of Theorem 4, we have that f=s/2, while from part (7), we have f+g=s. Then, B is a scalar matrix, while from parts (9) and (11), we get u=v=t/2. Thus, C is also a scalar matrix. Therefore, the representation is not faithful. Also, if ac≠0, then from parts (5) and (8), we get f=g, and from parts (9) and (12), we get u=v. Therefore, the representation is not faithful. Similarly, if ad≠0, then from parts (5) and (6), such that f=g, and from parts (9) and (10), we have u=v. Therefore, we must have a=0. In a similar argument, one must have d=0.
Now, from part (7), we have f+g=s, and in part (4), we get c2=rs/2−b2. Then, c=±rs/2−x2+y2, if we let b=x+iy.
A real representation can be obtained by takingy=0 in Theorem 6. A practically useful representation is when A is a nilpotent matrix, i.e., the case when x=y=0. So, we have the following corollary.Corollary 2.
Ifq=−1, then the −1-deformed Lie algebra t−1 is isomorphic to gl2,ℝ and has faithful representations of degree 2 as the least degree if and only if rs>0. The representation matrices of the generators K^4,K^3,K^1, and K^2 in (12), are, respectively, for real m≠t/2.(24)A=00±rs20,A†=0±rs200,B=s2I2,C=m00t−m.Assuming thatq∉−1,0 and using a similar technique as that of the proof of Theorems 5 and 6, we have the following theorem.Lemma 4.
Letq∉−1,0. In a 2-dimensional representation, the representation matrix of K^4 is a triangular nilpotent matrix. Its degree of nilpotency equals 2.Proof.
Ifab≠0, then from parts (5) and (7) of Theorem 4, we have f−qf=s=f−qg; hence, f=g as q≠0. And from parts (9) and (11), we have u−qu=u−qv, then u=v. Therefore, the representation matrices of K^1 and K^2 are scalar matrices, which are linearly dependent, and the representation is not faithful.
Ifac≠0, then from parts (5) and (8) of Theorem 4, we have f−qf=g−qf , which implies that f=g. Similarly, from parts (9) and (12), we conclude that u=v.
Thus, ifa≠0, we get b=c=0, which causes the representation matrix A, of K^4, to be diagonal as B and C. This leads to a representation which is not faithful.
Therefore, for a faithful representation oftq, if q≠0, then one mus havea=0. A similar argument shows that also one should have d=0.
Now, ifbc≠0, then from parts (7) and (8) of Theorem 4, we have f−qg=g−qf; hence, 1+qf−g=0 and since q≠−1, then f=g. And from parts (11) and (12), u−qv=v−qu, then 1+qu−v=0, then u=v.
Therefore,A has at most one nonzero element, which cannot be a diagonal element otherwise the representation is not faithful.
Now, we come to the general case, whereq∉−1,0.Theorem 7.
Forq∉−1,0, the q-deformed Lie algebra tq:K^4,K^3q=rK^1,K^1,K^4q=sK^4,K^3,K^1q=sK^3,K^2,K^4q=tK^4,K^3,K^2q=tK^3 has faithful matrix representations of degree 2, as the least degree, where the representation matrices are for x,m∈ℝ, such that x2≤rs/1+q2, and m≠−qt/1+q2, we have(25)K^4=0x±irs−1+q2x21+q200,K^3=K^4†=00x∓irs−1+q2x21+q20,K^1=s1+q2100−q,K^2=t+mq00m.Moreover,rs>0 is a necessary condition for the representation.Proof.
From Lemma4, let the representation matrices of K^4,K^3,K^1, and K^2, be, respectively, the 2×2 matrices A=x+iye12≠0,A†=x−iye21,B=diagf,g, and C=diagu,v, where f,g,u,v,x,y∈ℝ. So, in Theorem 4, let b=x+iy≠0, and a=c=d=0. Thus, from parts (2) and (3) of Theorem 4, we have that(26)f=x2+y2r,g=−qx2+y2r.
This shows thatg=−qf. From part (7) of Theorem 4, and by using (26), we get(27)s=1+q2f=1+q2x2+y2r.
Thus, we should havers>0.
Also, from (27), solving for y, we have(28)y2=rs−1+q2x21+q2≥0.
Therefore,(29)x2≤rs1+q2.
Now, from (21) and part (11) of Theorem 4, we have u−qv=t. Taking v=m, then u=t+mq. For the linear independence of B and C, it can be shown that m≠−qt/1+q2. For, if for some α≠0, we have m=−qαs/1+q2, then one should have αs/1+q2≠t+mq, which implies that m/−q≠t+mq.
Takingy=0, we obtain a real representation. In such a case, tq is, actually, the whole general linear algebra, gl2,ℝ.
As special cases, the obtained results agree with the previous cases whenq=1.Corollary 3.
The light amplifier model, namely,K^+,K^−=−2K^0,K^0,K^±=±K^± has no faithful matrix representation.Proof.
In the light amplifier model (6), since s=1 and r=−2, then rs<0, contradicting the necessary condition of Theorem 7.Corollary 4.
The two optical atom models, namely,K^+,K^−=K^0,K^0,K^±=±2K^± has the faithful matrix representation(30)K^+=0100,K^−=K^+†=0010,K^0=100−1.Proof.
In this model, we have,q=1,r=1, and s=2. Take x=1 and y=0.Corollary 5.
The representation matrices of the Tavis–Cummings model in (3)(31)K^4,K^3=K^1,K^1,K^4=2K^4,K^1,K^3=−2K^3,K^2,K^4=−K^4,K^2,K^3=K^3,K^1,K^2=0,K^4=0a±i1−a200,K^3=K^4†,K^1=100−1, and K^2=b00b+1, for a,b∈ℝ such that a2≤1, which generalize those representation matrices of the (TC) model which were given in [21], when taking a=1 and b=0.Proof.
In this model, we haveq=1,r=1,s=2, and t=−1. Take m=b+1.
We conclude with the following summary.


## 6. Summary

Our main purpose in this work is to find, if any, the possible faithful matrix representations of the least degree, satisfying physical conditions, fortq, given in (12), as a generalized model of t1, that is, when q=1, given in (3).It was recommended in Theorem2, that the commutation relation K^1,K^2=0 not to be included in (3), the defining relations of t1. Also, K^1,K^2q≠0 for q≠1 need not to be included in (12), the defining relations of tq.The faithful representations oftq, for the special cases, namely, q=0 and q=−1 are discussed in Theorems 5 and 6, respectively, where the field of consideration is ℂ. The real representations of both special cases are given in Corollarys 1 and 2, respectively.The faithful matrix representation oftq, for the general cases, are given for q∉−1,0 in Theorem 7. In Corollary 3, the real representation matrices for the generators of tq are given.



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*Source: 6737287-2022-09-19.xml*