On Faithful Matrix Representations of q-Deformed Models in Quantum Optics

Abstract

Consider the $q$-deformed Lie algebra, ${\mathfrak{t}}_q:{{\widehat{K}}_1,{\widehat{K}}_2}_q=1-q{\widehat{K}}_1{\widehat{K}}_2,{{\widehat{K}}_3,{\widehat{K}}_1}_q=s{\widehat{K}}_3$, ${{\widehat{K}}_1,{\widehat{K}}_4}_q=s{\widehat{K}}_4,{{\widehat{K}}_3,{\widehat{K}}_2}_q=t{\widehat{K}}_3,{{\widehat{K}}_2,{\widehat{K}}_4}_q=t{\widehat{K}}_4$, and ${{\widehat{K}}_4,{\widehat{K}}_3}_q=r{\widehat{K}}_1$, where $r,s,t\in ℝ-0$, subject to the physical properties: ${\widehat{K}}_1$ and ${\widehat{K}}_2$ are real diagonal operators, and ${\widehat{K}}_3={\widehat{K}}_4^{†}$, ($†$ is for Hermitian conjugation). The $q$-deformed Lie algebra, ${\mathfrak{t}}_q$ is introduced as a generalized model of the Tavis–Cummings model (Tavis and Cummings 1968, Bashir and Sebawe Abdalla 1995), namely, ${\widehat{K}}_1,{\widehat{K}}_2=0,{\widehat{K}}_1,{\widehat{K}}_3=-2{\widehat{K}}_3,{\widehat{K}}_1,{\widehat{K}}_4=2{\widehat{K}}_4,{\widehat{K}}_2,{\widehat{K}}_3={\widehat{K}}_3,{\widehat{K}}_2,{\widehat{K}}_4={\widehat{K}}_4$, and ${\widehat{K}}_4,{\widehat{K}}_3={\widehat{K}}_1$, which is subject to the physical properties ${\widehat{K}}_1$ and ${\widehat{K}}_2$ are real diagonal operators, and ${\widehat{K}}_3={\widehat{K}}_4^{†}$. Faithful matrix representations of the least degree of ${\mathfrak{t}}_q$ are discussed, and conditions are given to guarantee the existence of the faithful representations.

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1. Introduction

In the field of quantum optics, the Tavis–Cummings (TC) model [1] involves the multiphoton interaction of $N$$>1$ 2-level atoms with a single mode quantized field, in the absence of any radiation damping. It is a generalization of the Jaynes–Cummings (JC) model [2] of a single 2-level atom interacting via single photon transition with a single quantized field mode. The Lie algebra approach in [3–5] was employed in [6], where the Hamiltonian description of the the TC model of a coupled system of a boson (single mode of the radiation field) and $N$ fermions ($N$ two-level atoms) is given by $H={\omega }_0{\widehat{c}}^{†}\widehat{c}+1/2{\omega }_1{\widehat{J}}_z+g{\widehat{c}}^{†m}{\widehat{J}}_{-}+{\widehat{c}}^m{\widehat{J}}_{+}$, where $\widehat{c}$ and ${\widehat{c}}^{†}$ are the boson field annihilation and creation operators satisfying $\widehat{c},{\widehat{c}}^{†}=1$, while ${\widehat{J}}_{\pm }$ are the usual collective angular momentum operators, with ${\omega }_0$ and ${\omega }_1$ denoting the field and atomic frequencies, while $g$ and $m$ are the coupling constant and the number of photons, respectively. Using the transformation in [7–9], ${\widehat{c}}^m=f{\widehat{n}}_c\widehat{b}$, and ${{\widehat{c}}^{†}}^m={\widehat{b}}^{†}f{\widehat{n}}_c,$ the operators $\widehat{b}$ and ${\widehat{b}}^{†}$ are boson operators satisfying the commutation relation ${\widehat{b}}_i,{\widehat{b}}_j^{†}={\delta }_{ij}$, and the functions $f{\widehat{n}}_c$ are real functions of the number operator ${\widehat{n}}_c=\widehat{c}{\widehat{c}}^{†}$ given by $f{\widehat{n}}_c={{\widehat{n}}_c+m!/{\widehat{n}}_c!{\widehat{n}}_c/m+1!}^{1/2}$ , and on using the Holstien–Primakoff transformation [10], ${\widehat{J}}_{-}=\sqrt{\widehat{J}-{\widehat{n}}_a}\widehat{a}{e}^{i\varphi t},{\widehat{J}}_{+}={\widehat{a}}^{†}\sqrt{\widehat{J}-{\widehat{n}}_a}{e}^{-i\varphi t}$, and ${\widehat{n}}_a={\widehat{a}}^{†}\widehat{a}=1/2\widehat{J}+{\widehat{J}}_z$, where the phase $\varphi t$ is an arbitrary function of time, the Hamiltonian becomes$$\begin{array}{}\text{(1)}& H={\omega }_1{\widehat{a}}^{†}\widehat{a}+{\omega }_2{\widehat{b}}^{†}\widehat{b}+g{\widehat{b}}^{†}f{\widehat{n}}_c\widehat{a}\sqrt{\widehat{J}-{\widehat{n}}_a}{e}^{i\varphi t}+c.c.,\end{array}$$where ${\omega }_2=m{\omega }_0$. On approximating ${\widehat{n}}_c$ and ${\widehat{n}}_a$ by their $C$-number time-dependent functions $n_{c}t$ and $n_{a}t$, then the Hamiltonian (1) is bilinearized and has the form$$\begin{array}{}\text{(2)}& H={\omega }_1{\widehat{a}}^{†}\widehat{a}+{\omega }_2{\widehat{b}}^{†}\widehat{b}+\lambda t\widehat{a}{\widehat{b}}^{†}{e}^{i\varphi t}+{\widehat{a}}^{†}\widehat{b}{e}^{-i\varphi t},\end{array}$$where $\lambda t=g\sqrt{\widehat{J}-{\widehat{n}}_{a}t}f{\widehat{n}}_{c}t$, $\widehat{J}\ge {\widehat{n}}_a$. 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 ${\widehat{K}}_l,{\widehat{K}}_2,{\widehat{K}}_3,{\widehat{K}}_4$ as follows: ${\widehat{K}}_1={\widehat{a}}^{†}\widehat{a}-{\widehat{b}}^{†}\widehat{b},{\widehat{K}}_2={\widehat{b}}^{†}\widehat{b},{\widehat{K}}_3=\widehat{a}{\widehat{b}}^{†},{\widehat{K}}_4={\widehat{a}}^{†}\widehat{b}$, which satisfy the following commutation relations:$$\begin{array}{c}\text{(3)}& {\widehat{K}}_1,{\widehat{K}}_2=0,{\widehat{K}}_1,{\widehat{K}}_3=-2{\widehat{K}}_3,\\ {\widehat{K}}_1,{\widehat{K}}_4=2{\widehat{K}}_4,\\ {\widehat{K}}_2,{\widehat{K}}_3={\widehat{K}}_3,\\ {\widehat{K}}_2,{\widehat{K}}_4={\widehat{K}}_4,\\ {\widehat{K}}_4,{\widehat{K}}_3={\widehat{K}}_1.\end{array}$$

Then, the Hamiltonian (2) becomes$$\begin{array}{}\text{(4)}& H={\omega }_1{\widehat{K}}_1+{\omega }_1+{\omega }_2{\widehat{K}}_2+\lambda t{\widehat{K}}_3{e}^{i\varphi }+{\widehat{K}}_4{e}^{-i\varphi }.\end{array}$$

In [6], a matrix representation for the operators ${\widehat{K}}_1,{\widehat{K}}_2,{\widehat{K}}_3$ and ${\widehat{K}}_4$ satisfying (3) is found to be ${\widehat{K}}_1=\begin{array}{cc}1& 0\\ 0& -1\end{array},{\widehat{K}}_2=\begin{array}{cc}0& 0\\ 0& 1\end{array},{\widehat{K}}_3=\begin{array}{cc}0& 0\\ 1& 0\end{array},{\widehat{K}}_4=\begin{array}{cc}0& 1\\ 0& 0\end{array}$.

In [16], two optical atom models were introduced, namely, the model of the two-level optical atom as follows:$$\begin{array}{c}\text{(5)}& {\widehat{K}}_{+},{\widehat{K}}_{-}={\widehat{K}}_0,{\widehat{K}}_0,{\widehat{K}}_{\pm }=\pm 2{\widehat{K}}_{\pm },\end{array}$$and the model of a light amplifier, which corresponds to the $\mathfrak{s}\mathfrak{u}\mathrm{1,1}$ algebra$$\begin{array}{c}\text{(6)}& {\widehat{K}}_{+},{\widehat{K}}_{-}=-2{\widehat{K}}_0,{\widehat{K}}_0,{\widehat{K}}_{\pm }=\pm {\widehat{K}}_{\pm }.\end{array}$$

The Hamiltonian of these two coupled quantized harmonic oscillators models is of the following form:$$\begin{array}{}\text{(7)}& H=\omega {\widehat{a}}^{†}\widehat{a}-{\widehat{b}}^{†}\widehat{b}+\lambda T{\widehat{a}}^{†}\widehat{b}+\widehat{a}{\widehat{b}}^{†},\end{array}$$where $\lambda T$ is an arbitrary time-dependent coupling parameter, and the Boson operators of the two-level optical atom obey the commutation relations:$$\begin{array}{}\text{(8)}& \widehat{a},{\widehat{a}}^{†}=1=\widehat{b},{\widehat{b}}^{†}.\end{array}$$

As a combination of the three operators, namely, ${\widehat{K}}_0,{\widehat{K}}_{\pm }$, the Hamiltonian can be written as$$\begin{array}{}\text{(9)}& H=\omega {\widehat{K}}_0+\lambda T{\widehat{K}}_{+}+{\widehat{K}}_{-},\end{array}$$where ${\widehat{K}}_0={\widehat{a}}^{†}\widehat{a}-{\widehat{b}}^{†}\widehat{b}$, and ${\widehat{K}}_{+}={\widehat{a}}^{†}\widehat{b}={\widehat{K}}_{-}^{†}$, with ${\widehat{K}}_{+}+{\widehat{K}}_{-}$ 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, ${\widehat{K}}_0,{\widehat{K}}_{\pm }$ is the $\mathfrak{s}\mathfrak{u}2$ algebra. In [18, 19], faithful matrix representations of the least degree for the two-level optical atom were found to be$$\begin{array}{c}\text{(10)}& {\widehat{K}}_{+}=\begin{array}{cc}0& 1\\ 0& 0\end{array},{\widehat{K}}_{-}={\widehat{K}}_{+}^{†}=\begin{array}{cc}0& 0\\ 1& 0\end{array},\\ {\widehat{K}}_0=\begin{array}{cc}1& 0\\ 0& -1\end{array},\end{array}$$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 $\mathrm{0,1}$, for the particular model of Fermion oscillators.

Definition 1.

Let $X$ and $Y$ be $n\times n$ matrices. The $q$-deformed Lie bracket of $X$ and $Y$ is defined to be$$\begin{array}{}\text{(11)}& {X,Y}_q=XY-qYX,\hspace{1em}\mathrm{for}q\in ℝ.\end{array}$$

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), ${\widehat{K}}_1,{\widehat{K}}_4,$ and ${\widehat{K}}_3$ satisfy the same commutation relations of ${\widehat{K}}_0,{\widehat{K}}_{+},$ and ${\widehat{K}}_{-}$, 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, the $q$-deformed Lie algebra, ${\mathfrak{t}}_q$ is introduced as a generalization of the Tavis–Cummings model in (3), namely,$$\begin{array}{c}\text{(12)}& {\mathfrak{t}}_q:{{\widehat{K}}_1,{\widehat{K}}_2}_q=1-q{\widehat{K}}_1{\widehat{K}}_2,{{\widehat{K}}_3,{\widehat{K}}_1}_q=s{\widehat{K}}_3,\\ {{\widehat{K}}_1,{\widehat{K}}_4}_q=s{\widehat{K}}_4,\\ {{\widehat{K}}_3,{\widehat{K}}_2}_q=t{\widehat{K}}_3,\\ {{\widehat{K}}_2,{\widehat{K}}_4}_q=t{\widehat{K}}_4,\\ {{\widehat{K}}_4,{\widehat{K}}_3}_q=r{\widehat{K}}_1,\end{array}$$subject to the physical requirements, ${\widehat{K}}_3={\widehat{K}}_4^{†}$, and ${\widehat{K}}_1$ and ${\widehat{K}}_2$ are real diagonal operators.

The main purpose of this paper is to construct matrices of the least degree faithfully representing the $q$-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 ${\widehat{K}}_2$. Throughout this paper, $A,A^{†},B$, and $C$ are the representation matrices of ${\widehat{K}}_4,{\widehat{K}}_3,{\widehat{K}}_1,$ and ${\widehat{K}}_2$, respectively, of the model (12), with $B$ and $C$ as real diagonal matrices. On the other hand, $A,A^{†}$, and $B$ can represent ${\widehat{K}}_{+},{\widehat{K}}_{-}$, and ${\widehat{K}}_0$, 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 matrices $A,A^{†},B$, and $C$, satisfying$$\begin{array}{c}\text{(13)}& {\mathfrak{t}}_q:{B,C}_q=1-qBC,{A^{†},B}_q=s{A}^{†},\\ {B,A}_q=sA,\\ {A^{†},C}_q=t{A}^{†},\\ {C,A}_q=tA,\\ {A,A^{†}}_q=rB.\end{array}$$

Faithful matrix representations of degree 2 were proved to be the least degree. Also, ${\mathfrak{t}}_q$ was given as an extension of the generalized model of the $q$-deformed couple quantized harmonic oscillator by the operator ${\widehat{K}}_2$.

The representation matrices, for ${\mathfrak{t}}_q$, of degree 2 as the least degree, which satisfy the physical conditions, are given in Theorem 5 as follows.

For $q\notin -\mathrm{1,0}$, and $a,b\in ℝ$, such that $a^2\le rs/1+q^2$, we have$$\begin{array}{c}\text{(14)}& A=\begin{array}{cc}0& a\pm i\sqrt{\frac{rs-1+q^2{a}^2}{1+q^2}}\\ 0& 0\end{array},A^{†}=\begin{array}{cc}0& 0\\ a\mp i\sqrt{\frac{rs-1+q^2{a}^2}{1+q^2}}& 0\end{array},\\ B=\frac{s}{1+q^2}\begin{array}{cc}1& 0\\ 0& -q\end{array},\\ C=\begin{array}{cc}t+bq& 0\\ 0& b\end{array}.\end{array}$$

The special cases where $q=0$ and $q=-1$ are given in Theorems 5 and 6, respectively.

2. Preliminaries

Let $X$ and $Y$ be $n\times n$ matrices. The $q$-deformed Lie bracket of $X$ and $Y$ is defined, [11] to be ${X,Y}_q=XY-qYX$, for $q\in ℝ$. 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 ${X,Y}_1$ as $X,Y$, whenever needed. The matrix representation of the TC model were given for $q=1$ in [24]. So, the cases where $q\in ℝ-\mathrm{0,1}$ will always be in our interest.

Unless otherwise stated, the square matrices $A,A^{†},B,$ and $C$ denote the representation matrices of the operators ${\widehat{K}}_4,{\widehat{K}}_3,{\widehat{K}}_1,$ and ${\widehat{K}}_2$, respectively, of the model (12), and $\mathbf{0}$ is the zero square matrix of an appropriate size. The set ${\mathbb{N}}_n$ denotes the set $\mathrm{1,2},\dots n$ for positive integer $n$. All representations are supposed to satisfy the physical requirements that ${\widehat{K}}_1$ and ${\widehat{K}}_2$ are real diagonal operators. Also, $r,s,t\in ℝ$ such that $rst\ne 0$.

3. Basic Properties of the $q$-Deformed Lie Bracket

For $i,j\in {\mathbb{N}}_n$, let $e_{ij}$ be the matrix unit, which is an $n\times n$ matrix whose element in the $i^{\text{th}}$ row and $j^{\text{th}}$ column is $1$ and zero elsewhere. In other words, the $e_{ij}^{,}s$ are the elements of the standard basis of $M_{n\times n}ℂ$, the set of all $n\times n$ matrices over the field $ℂ$ of complex numbers. Since$$\begin{array}{}\text{(15)}& e_{ij}{e}_{kl}={\delta }_{jk}{e}_{il},\end{array}$$where ${\delta }_{ij}$ is the Kronecker delta, we have the following lemma.

Lemma 1.

Consider the matrix units, $e_{ij}$ for $i,j\in {\mathbb{N}}_n$, then for $q\in ℝ$, we have$$\begin{array}{}\text{(16)}& {e_{ij},e_{kl}}_q={\delta }_{jk}{e}_{il}-q{\delta }_{il}{e}_{kj}.\end{array}$$

(1) If $j\ne k$ and $l\ne i$, then ${e_{ij},e_{kl}}_q=\mathbf{0}$.

(2) If $i\ne j$, then ${e_{ij},e_{ij}}_q=\mathbf{0}$.

(3) If $l\ne i$, then ${e_{ij},e_{jl}}_q=e_{il}$.

(4) If $j\ne k$, then ${e_{ij},e_{ki}}_q=-q{e}_{kj}$.

(5) If $i\ne j$, then ${e_{ij},e_{ji}}_q=e_{ii}-q{e}_{jj}$.

(6) ${e_{ii},e_{ii}}_q=1-q{e}_{ii}$.

Proof.

From (11) and (15), we have ${e_{ij},e_{kl}}_q=e_{ij}{e}_{kl}-q{e}_{kl}{e}_{ij}={\delta }_{jk}{e}_{il}-q{\delta }_{il}{e}_{kj}$. Other parts are easily driven from (16).

To be self-contained, we copy the following properties of the $q$-deformed Lie bracket from [25] in Theorem 1. We also, add other properties in Lemma 2.

Theorem 1.

Let $M_{n\times n}ℂ$ be the set of all $n\times n$ matrices over a field $ℂ$ and $q\in ℝ$. Let $\alpha ,\beta \in ℂ$. Then, for $X,Y$ and $Z$ in $M_{n\times n}ℂ$, we have

(1) $tr{X,Y}_q=tr{Y,X}_q$.

(2) ${X,Y}_q^{†}={Y^{†},X^{†}}_q$.

(3) ${X+Y,Z}_q={X,Z}_q+{Y,Z}_q$.

(4) ${X,Y+Z}_q={X,Y}_q+{X,Z}_q$.

(5) ${X,Y}_q=qX,Y+1-qXY$.

(6) ${\alpha A,\beta B}_q=\alpha \beta {A,B}_q$.

(7) ${X,Y}_q=X,Y+1-qYX$.

(8) ${X,X}_q=1-q{X}^2$.

(9) ${Y,X}_q={X,Y}_q-1+qX,Y$.

(10) ${X,{Y,Z}_q}_q+{Y,{Z,X}_q}_q+{Z,{X,Y}_q}_q=1-qXYZ+YZX+ZXY-qXZY+YXZ+ZYX.$

Part 10 is the $q$-deformed version of the Jacobi identity. So, we add here the following properties.

Lemma 2.

Let $M_{n\times n}ℂ$ be the set of all $n\times n$ matrices over $ℂ$ and $q\in ℝ$. Let $\alpha ,\beta \in ℂ$. Then, for $X,Y$, and $Z$ in $M_{n\times n}ℂ$, we have

(1) ${XY,Z}_q=XY,Z+1-qZXY=1-qXYZ+qXY,Z$ and also,

(2) ${XY,Z}_q=X,YZ+{Y,ZX}_q$.

(3) ${X,YZ}_q=X,YZ+1-qYZX=1-qXYZ+qX,YZ$ and also,

(4) ${X,YZ}_q={XY,Z}_q+qZX,Y$.

(5) ${{X,Y}_q,Z}_q+{{Y,Z}_q,X}_q+{{Z,X}_q,Y}_q=1-qXYZ+YZX+ZXY-qXZY+YXZ+ZYX$.

Proof.

For part (1), we have ${XY,Z}_q=XYZ-qZXY=XYZ+ZXY-ZXY-qZXY=XY,Z+1-qZXY$. Also, ${XY,Z}_q=XYZ-qZXY=XYZ-qXYZ+qXYZ-qZXY=1-qXYZ+qXY,Z$. For part (2), ${XY,Z}_q=XYZ-qZXY=XYZ-qYZX+qYZX-qZXY={X,YZ}_q+qY,ZX$.

${X,YZ}_q=XYZ-qYZX=XYZ-YZX+YZX-qYZX=X,YZ+1-qYZX$. 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 that ${\mathfrak{t}}_q$ has a faithful matrix representation of degree $n$, where the $n\times n$ matrices $A,A^{†},B,$ and $C$ are representation matrices of ${\widehat{K}}_4,{\widehat{K}}_3,{\widehat{K}}_1,$ and ${\widehat{K}}_2$, respectively, of the model (12). Thus, $A,A^{†},B,$ and $C$ are linearly independent with $B$ and $C$ as real diagonal matrices.

Lemma 3.

Let $u,v\in {\mathbb{N}}_n$, and $\alpha =u,v$ be a transposition. The representation obtained by applying $\alpha$ to the rows as well as to the columns of matrices $A,A^{†},B,$ and $C$ is a conjugate representation for ${\mathfrak{t}}_q$ and satisfies the physical requirements.

Proof.

Let $X$ be the elementary matrix obtained by applying $\alpha$ to the rows of $I_n$. Since $X=X^{-1}=X^T=X^{†}={X^{-1}}^{†}$, then $A^{\prime }=X^{-1}AX,{A^{\prime }}^{†}={X^{-1}AX}^{†}=X^{†}{A}^{†}{X^{-1}}^{†}=X^{-1}{A}^{†}X,B^{\prime }=X^{-1}BX$, and $C^{\prime }=X^{-1}CX$ are representation matrices for ${\widehat{K}}_4,{\widehat{K}}_3,{\widehat{K}}_1$, and ${\widehat{K}}_2$, respectively, of the model (12), satisfying $\begin{array}{c}{A^{\prime },{A^{\prime }}^{†}}_q={X^{-1}AX,X^{-1}{A}^{†}X}_q=X^{-1}AX{X}^{-1}{A}^{†}X-q{X}^{-1}{A}^{†}X{X}^{-1}AX=X^{-1}A{A}^{†}-q{A}^{†}A\\ X=X^{-1}{A,A^{†}}_{q}X=X^{-1}rBX=r{B}^{\prime }\end{array}$. Similarly, ${B^{\prime },A^{\prime }}_q=X^{-1}{B,A}_{q}X=X^{-1}sAX=s{A}^{\prime }$ and ${C^{\prime },A^{\prime }}_q=t{A}^{\prime }$. Also, ${B^{\prime },C^{\prime }}_q=X^{-1}{B,C}_{q}X$.

As ${\mathfrak{t}}_q$ is defined to be a generalization of the Tavis–Cummings model in (3), it should be pointed out that the $q$-deformed commutator ${{\widehat{K}}_1,{\widehat{K}}_2}_q$ cannot be equal to $\mathbf{0}$ for $q\ne 1$.

Theorem 2.

The matrix product $BC\ne \mathbf{0}.$ Moreover, the commutator ${B,C}_q=\mathbf{0}$ if $q=1$.

Proof.

Let $A,B,$ and $C$ be representation matrices of ${\widehat{K}}_4,{\widehat{K}}_1,$ and ${\widehat{K}}_2$, respectively, in (3). Suppose that ${B,C}_q=\mathbf{0}$, and $q\ne 1$. Since $B$ and $C$ are diagonal, then ${B,C}_q=1-qBC=\mathbf{0}$ if $BC=\mathbf{0}$. Thus, if $B_{ii}\ne 0$, then $C_{ii}=0$ for $i\in {\mathbb{N}}_n$. Using Lemma 3, to rearrange $B$ such that, $B=\text{diag}T,V$, where $T$ is an $l\times l$ nonsingular diagonal matrix, and $V={\mathbf{0}}_{n-l\times n-l}$ matrix, while $C=\text{diag}{\mathbf{0}}_{l\times l},S_{n-l\times n-l}$, for some $l\in {\mathbb{N}}_n$.

Since the diagonal elements of $B$ are all zeros for each $i=l+1,l+2,\dots ,n$; we have ${sA}_{ii}={{B,A}_q}_{ii}={BA}_{ii}-q{AB}_{ii}=0-q0=0$. So, $A_{ii}=0$ for $i>l$. Similarly, since for each $i=1,\dots ,l$; we have ${C,A}_{ii}={CA-qAC}_{ii}=0={tA}_{ii}$. Hence, $A_{ii}=0$ for $i\le l$. Thus, all diagonal elements of $A$ are zeros.

Since $A\ne \mathbf{0},$ otherwise the representation is not faithful, then for some $i\ne j$, there is an $A_{ij}\ne 0$. If $i>l$, then the $i^{\text{th}}$ row and the $i^{\text{th}}$ column of $B$ are zero row and zero column, respectively. As ${B,A}_q=sA,$ then${{B,A}_q}_{ij}={\displaystyle {\sum }_{t=1}^n{B}_{it}{A}_{tj}-q{A}_{it}{B}_{tj}}=0$, contradicting that ${sA}_{ij}\ne 0$, as $s\ne 0$. Similarly, if $i\le l$, then the $i^{\text{th}}$ row and the $i^{\text{th}}$ column of $C$ are zero row and zero column, respectively, and as ${C,A}_q=tA$, then ${{C,A}_q}_{ij}=0$ contradicting that ${tA}_{ij}\ne 0$, as $t\ne 0$.

As $B$ and $C$ are diagonal matrices, then their commutator ${B,C}_q=1-qBC$ is too trivial to be included in the defining relations of ${\mathfrak{t}}_q$. Then, we have the following theorem.

Theorem 3.

The defining relations of ${\mathfrak{t}}_q$ can be either$$\begin{array}{c}\text{(17)}& {\mathfrak{t}}_q:{A^{†},B}_q=s{A}^{†},{B,A}_q=sA,\\ {A^{†},C}_q=t{A}^{†},\\ {C,A}_q=tA,\\ {A,A^{†}}_q=rB,\end{array}$$or$$\begin{array}{c}\text{(18)}& {\mathfrak{t}}_q:{A,A^{†}}_q=rB,{B,A}_q=sA,\\ {C,A}_q=tA.\end{array}$$

Subject to the physical conditions: $B$ and $C$ are real diagonal operators.

Proof.

From part 2 of Theorem 1, we have ${{B,A}_q}^{†}={sA}^{†}$, which implies that ${A^{†},B}_q=s{A}^{†}$. Similarly, taking the Hermitian conjugate of both sides of the commutator, ${C,A}_q=tA$, we get ${A^{†},C}_q=t{A}^{†}$.

5. Least Degree Faithful Matrix Representation of the $q$-Deformed Tavis–Cummings Model

We first try to investigate whether ${\mathfrak{t}}_q$ has faithful representations of degree 2. In this section, unless otherwise stated, let $A=a{e}_{11}+b{e}_{12}+c{e}_{21}+d{e}_{22}$, $A^{†},B=f{e}_{11}+g{e}_{22},$ and $C=u{e}_{11}+v{e}_{22}$ be representation matrices of ${\widehat{K}}_4,{\widehat{K}}_3,{\widehat{K}}_1,$ and ${\widehat{K}}_2$ respectively in (12), where $a,b,c,d\in ℂ$ and $f,g,u,v\in ℝ$, satisfying that for $r,s,t\in ℝ-0$,$$\begin{array}{}\text{(19)}& {A,A^{†}}_q=rB,\text{(20)}& {B,A}_q=sA,\text{(21)}& {C,A}_q=tA,\end{array}$$where the matrix units $e_{ij}$’s are $2\times 2$ matrices for $i,j\in {\mathbb{N}}_2$.

Theorem 4.

Consider the $2\times 2$ matrix units, $e_{11},e_{12},e_{21},$ and $e_{22}$. Let $A=a{e}_{11}+b{e}_{12}+c{e}_{21}+d{e}_{22}$, $A^{†},B=f{e}_{11}+g{e}_{22},$ and $C=u{e}_{11}+v{e}_{22}$ be representation matrices of ${\widehat{K}}_4,{\widehat{K}}_3,{\widehat{K}}_1,$ and ${\widehat{K}}_2$, respectively, where $a,b,c,d\in ℂ$ and $f,g,u,v\in ℝ$. Then, for $r,s,t\in ℝ-0$, we have

(1) $a\overline{c}+b\overline{d}-qb\overline{a}+d\overline{c}=0$,

(2) $1-q{a}^2+b^2-q{c}^2=rf$,

(3) $1-q{d}^2+c^2-q{b}^2=rg$,

(4) $trB=f+g=1-q/r{a}^2+b^2+c^2+d^2,$

(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 Theorem 4, in order to build the representation matrices for ${\mathfrak{t}}_q$ 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.

If $q=0$, then the 0-deformed Lie algebra ${\mathfrak{t}}_0$ has faithful representations of degree 2, as the least degree. The representation matrices of the generators ${\widehat{K}}_4,{\widehat{K}}_3,{\widehat{K}}_1,$ and ${\widehat{K}}_2$ are, respectively, for $x,y,m\in ℝ$, such that $0\le y^2\le rs-x^2$ and $rstm\ne 0$,

(1) $A=\begin{array}{cc}x+iy& \pm \sqrt{rs-x^2+y^2}\\ 0& 0\end{array},A^{†}=\begin{array}{cc}x-iy& 0\\ \pm \sqrt{rs-x^2+y^2}& 0\end{array},B=\begin{array}{cc}s& 0\\ 0& 0\end{array}$, and $C=\begin{array}{cc}t& 0\\ 0& m\end{array}$, or

(2) $A=\begin{array}{cc}\pm \sqrt{rs-x^2+y^2}& x+iy\\ 0& 0\end{array},A^{†}=\begin{array}{cc}\pm \sqrt{rs-x^2+y^2}& 0\\ x-iy& 0\end{array},B=\begin{array}{cc}s& 0\\ 0& 0\end{array}$, and $C=\begin{array}{cc}t& 0\\ 0& m\end{array}$.

Proof.

Let $q=0$. If $ab\ne 0$, then from part (5) in Theorem 4, we have $f=s$, then we get from part (2) that $b^2=rs-a^2$. Thus, if $a=x+iy$, then $b^2=rs-x^2+y^2$. Thus, $b=\pm \sqrt{rs-x^2+y^2}$. For faithful representation, take $m\ne 0$. Cases, if $ac\ne 0,ad\ne 0,bc\ne 0,$ or $bd\ne 0$ lead to representations where ${\widehat{K}}_1$ and ${\widehat{K}}_2$ are represented by scalar matrices, and hence, the representations are not faithful. If $cd\ne 0$, then the representation is conjugate to either of the given representations.

A real representation can be obtained by taking $y=0$ 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.

If $q=0$, then the 0-deformed Lie algebra ${\mathfrak{t}}_0$ is isomorphic to $\mathfrak{g}\mathfrak{l}2,ℝ$ and has faithful representations of degree 2 as the least degree if and only if $rs>0$. The representation matrices of the generators ${\widehat{K}}_4,{\widehat{K}}_3,{\widehat{K}}_1,$ and ${\widehat{K}}_2$ in (12), are, respectively, for any $m\in ℝ-0$,$$\begin{array}{c}\text{(22)}& A=\begin{array}{cc}0& \pm \sqrt{rs}\\ 0& 0\end{array},A^{†}=\begin{array}{cc}0& 0\\ \pm \sqrt{rs}& 0\end{array},\\ B=\begin{array}{cc}s& 0\\ 0& 0\end{array},\\ C=\begin{array}{cc}t& 0\\ 0& m\end{array}.\end{array}$$

Theorem 6.

If $q=-1$, then the $-1$-deformed Lie algebra ${\mathfrak{t}}_{-1}$ has faithful representations of degree 2 as the least degree. The representation matrices of the generators ${\widehat{K}}_4,{\widehat{K}}_3,{\widehat{K}}_1,$ and ${\widehat{K}}_2$ in (12), are, respectively,$$\begin{array}{c}\text{(23)}& A=\begin{array}{cc}0& x+iy\\ \pm \sqrt{\frac{rs}{2}-x^2+y^2}& 0\end{array},A^{†}=\begin{array}{cc}0& \pm \sqrt{\frac{rs}{2}-x^2+y^2}\\ x-iy& 0\end{array},\\ B=\frac{s}{2}{I}_2,\\ C=\begin{array}{cc}m& 0\\ 0& t-m\end{array},\end{array}$$where $x,y\in ℝ$, and $m\ne t/2$ such that $x^2+y^2\le rs/2$.

Proof.

Let $q=-1$. We claim that we should have $a=0$. The case where $a\ne 0$ and $b=c=0$ leads to the representation matrix $A$ is diagonal and then, the representation is not faithful.

If $ab\ne 0$, 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 $ac\ne 0$, 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 $ad\ne 0$, 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 $c^2=rs/2-b^2$. Then, $c=\pm \sqrt{rs/2-x^2+y^2}$, if we let $b=x+iy$.

A real representation can be obtained by taking $y=0$ 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.

If $q=-1$, then the $-1$-deformed Lie algebra ${\mathfrak{t}}_{-1}$ is isomorphic to $\mathfrak{g}\mathfrak{l}2,ℝ$ and has faithful representations of degree 2 as the least degree if and only if $rs>0$. The representation matrices of the generators ${\widehat{K}}_4,{\widehat{K}}_3,{\widehat{K}}_1,$ and ${\widehat{K}}_2$ in (12), are, respectively, for real $m\ne t/2$.$$\begin{array}{c}\text{(24)}& A=\begin{array}{cc}0& 0\\ \pm \sqrt{\frac{rs}{2}}& 0\end{array},A^{†}=\begin{array}{cc}0& \pm \sqrt{\frac{rs}{2}}\\ 0& 0\end{array},\\ B=\frac{s}{2}{I}_2,\\ C=\begin{array}{cc}m& 0\\ 0& t-m\end{array}.\end{array}$$

Assuming that $q\notin -\mathrm{1,0}$ and using a similar technique as that of the proof of Theorems 5 and 6, we have the following theorem.

Lemma 4.

Let $q\notin -\mathrm{1,0}$. In a 2-dimensional representation, the representation matrix of ${\widehat{K}}_4$ is a triangular nilpotent matrix. Its degree of nilpotency equals 2.

Proof.

If $ab\ne 0$, then from parts (5) and (7) of Theorem 4, we have $f-qf=s=f-qg$; hence, $f=g$ as $q\ne 0$. And from parts (9) and (11), we have $u-qu=u-qv$, then $u=v$. Therefore, the representation matrices of ${\widehat{K}}_1$ and ${\widehat{K}}_2$ are scalar matrices, which are linearly dependent, and the representation is not faithful.

If $ac\ne 0$, 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, if $a\ne 0$, we get $b=c=0$, which causes the representation matrix $A$, of ${\widehat{K}}_4$, to be diagonal as $B$ and $C$. This leads to a representation which is not faithful.

Therefore, for a faithful representation of ${\mathfrak{t}}_q$, if $q\ne 0$, then one mus have$a=0$. A similar argument shows that also one should have $d=0$.

Now, if $bc\ne 0$, then from parts (7) and (8) of Theorem 4, we have $f-qg=g-qf$; hence, $1+qf-g=0$ and since $q\ne -1$, 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, where $q\notin -\mathrm{1,0}$.

Theorem 7.

For $q\notin -\mathrm{1,0}$, the $q$-deformed Lie algebra ${\mathfrak{t}}_q:{{\widehat{K}}_4,{\widehat{K}}_3}_q=r{\widehat{K}}_1,{{\widehat{K}}_1,{\widehat{K}}_4}_q=s{\widehat{K}}_4,{{\widehat{K}}_3,{\widehat{K}}_1}_q=s{\widehat{K}}_3,{{\widehat{K}}_2,{\widehat{K}}_4}_q=t{\widehat{K}}_4,{{\widehat{K}}_3,{\widehat{K}}_2}_q=t{\widehat{K}}_3$ has faithful matrix representations of degree 2, as the least degree, where the representation matrices are for $x,m\in ℝ$, such that $x^2\le rs/1+q^2$, and $m\ne -qt/1+q^2$, we have$$\begin{array}{c}\text{(25)}& {\widehat{K}}_4=\begin{array}{cc}0& x\pm i\sqrt{\frac{rs-1+q^2{x}^2}{1+q^2}}\\ 0& 0\end{array},{\widehat{K}}_3={\widehat{K}}_4^{†}=\begin{array}{cc}0& 0\\ x\mp i\sqrt{\frac{rs-1+q^2{x}^2}{1+q^2}}& 0\end{array},\\ {\widehat{K}}_1=\frac{s}{1+q^2}\begin{array}{cc}1& 0\\ 0& -q\end{array},\\ {\widehat{K}}_2=\begin{array}{cc}t+mq& 0\\ 0& m\end{array}.\end{array}$$

Moreover, $rs>0$ is a necessary condition for the representation.

Proof.

From Lemma 4, let the representation matrices of ${\widehat{K}}_4,{\widehat{K}}_3,{\widehat{K}}_1,$ and ${\widehat{K}}_2$, be, respectively, the $2\times 2$ matrices $A=x+iy{e}_{12}\ne \mathbf{0},A^{†}=x-iy{e}_{21},B=\text{diag}f,g,$ and $C=\text{diag}u,v$, where $f,g,u,v,x,y\in ℝ$. So, in Theorem 4, let $b=x+iy\ne 0$, and $a=c=d=0$. Thus, from parts (2) and (3) of Theorem 4, we have that$$\begin{array}{c}\text{(26)}& f=\frac{x^2+y^2}{r},g=-\frac{q{x}^2+y^2}{r}.\end{array}$$

This shows that $g=-qf$. From part (7) of Theorem 4, and by using (26), we get$$\begin{array}{}\text{(27)}& s=1+q^{2}f=\frac{1+q^2{x}^2+y^2}{r}.\end{array}$$

Thus, we should have $rs>0$.

Also, from (27), solving for $y$, we have$$\begin{array}{}\text{(28)}& y^2=\frac{rs-1+q^2{x}^2}{1+q^2}\ge 0.\end{array}$$

Therefore,$$\begin{array}{}\text{(29)}& x^2\le \frac{rs}{1+q^2}.\end{array}$$

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 $m\ne -qt/1+q^2$. For, if for some $\alpha \ne 0$, we have $m=-q\alpha s/1+q^2$, then one should have $\alpha s/1+q^2\ne t+mq$, which implies that $m/-q\ne t+mq$.

Taking $y=0$, we obtain a real representation. In such a case, ${\mathfrak{t}}_q$ is, actually, the whole general linear algebra, $\mathfrak{g}\mathfrak{l}2,ℝ$.

As special cases, the obtained results agree with the previous cases when $q=1$.

Corollary 3.

The light amplifier model, namely, ${\widehat{K}}_{+},{\widehat{K}}_{-}=-2{\widehat{K}}_0,\hspace{1em}{\widehat{K}}_0,{\widehat{K}}_{\pm }=\pm {\widehat{K}}_{\pm }$ 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, ${\widehat{K}}_{+},{\widehat{K}}_{-}={\widehat{K}}_0,{\widehat{K}}_0,{\widehat{K}}_{\pm }=\pm 2{\widehat{K}}_{\pm }$ has the faithful matrix representation$$\begin{array}{c}\text{(30)}& {\widehat{K}}_{+}=\begin{array}{cc}0& 1\\ 0& 0\end{array},{\widehat{K}}_{-}={\widehat{K}}_{+}^{†}=\begin{array}{cc}0& 0\\ 1& 0\end{array},\\ {\widehat{K}}_0=\begin{array}{cc}1& 0\\ 0& -1\end{array}.\end{array}$$

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)$$\begin{array}{c}\text{(31)}& {\widehat{K}}_4,{\widehat{K}}_3={\widehat{K}}_1,{\widehat{K}}_1,{\widehat{K}}_4=2{\widehat{K}}_4,\\ {\widehat{K}}_1,{\widehat{K}}_3=-2{\widehat{K}}_3,\\ {\widehat{K}}_2,{\widehat{K}}_4=-{\widehat{K}}_4,\\ {\widehat{K}}_2,{\widehat{K}}_3={\widehat{K}}_3,\\ {\widehat{K}}_1,{\widehat{K}}_2=0,\end{array}$$${\widehat{K}}_4=\begin{array}{cc}0& a\pm i\sqrt{1-a^2}\\ 0& 0\end{array},{\widehat{K}}_3={\widehat{K}}_4^{†},{\widehat{K}}_1=\begin{array}{cc}1& 0\\ 0& -1\end{array},$ and ${\widehat{K}}_2=\begin{array}{cc}b& 0\\ 0& b+1\end{array}$, for $a,b\in ℝ$ such that $a^2\le 1$, 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 have $q=1,r=1,s=2,$ 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, for ${\mathfrak{t}}_q$, given in (12), as a generalized model of ${\mathfrak{t}}_1$, that is, when $q=1$, given in (3).

It was recommended in Theorem 2, that the commutation relation ${\widehat{K}}_1,{\widehat{K}}_2=\mathbf{0}$ not to be included in (3), the defining relations of ${\mathfrak{t}}_1$. Also, ${{\widehat{K}}_1,{\widehat{K}}_2}_q\ne \mathbf{0}$ for $q\ne 1$ need not to be included in (12), the defining relations of ${\mathfrak{t}}_q$.

The faithful representations of ${\mathfrak{t}}_q$, 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 of ${\mathfrak{t}}_q$, for the general cases, are given for $q\notin -\mathrm{1,0}$ in Theorem 7. In Corollary 3, the real representation matrices for the generators of ${\mathfrak{t}}_q$ are given.

Acknowledgments

The authors are grateful to Prof. S. S. Hassan of the University of Bahrain for fruitful discussions and reading the manuscript. The undeniable support of Kuwait University is very much appreciated.

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