OF A GENERALIZED TAVIS-CUMMINGS MODEL
L. A. M. HANNA
Received 13 February 2002 and in revised form 9 July 2002
Consider the Lie algebrasLsr,t:[K1, K2] =sK3,[K3, K1] =rK1,[K3, K2] =
−rK2, [K3, K4] =0, [K4, K1] =−tK1, and [K4, K2] =tK2, subject to the physical conditions, K3 and K4 are real diagonal operators represent- ing energy, K2 =K1†, and the Hamiltonian H=ω1K3+ (ω1+ω2)K4+ λ(t)(K1e−iφ+K2eiφ)is a Hermitian operator. Matrix representations are discussed and faithful representations of least degree forLsr,tsatisfying the physical requirements are given for appropriate values ofr, s, t∈R.
1. Introduction
Introducing an algebraic method to solve certain types of linear par- tial differential equations, Steinberg[6]exploited the Lie-algebraic de- composition formulas of Baker, Campbell, Hausdorff, and Zassenhaus (cf.[7])and their matrix realization. A faithful matrix representation of low degree is required. In [2,3,4], the faithful matrix representations of least degree were discussed for the Lie algebraLsr generated byK+, K−, andK0 satisfying the commutation relations:[K0, K±] =±rK± and [K+, K−] =sK0subject to the physical propertiesK−=K+† (†for Hermit- ian conjugation), K0 is a real diagonal operator, and(K++K−) is real.
The Lie algebraLsr was introduced as a generalization of the coupled quantized harmonic oscillators[5]namely, the model of light amplifier L−21 , and the model of two-level optical atom L21, whose Hamiltonian modelH=K0+λ(K++K−),λis the coupling parameter. Note that,L12is exactly the Lie algebrasl(2).
In this paper,Lsr,tis considered to be the Lie algebra generated byK1, K2, K3, and K4, satisfying the commutation relations: [K1, K2] =sK3,
Copyrightc2003 Hindawi Publishing Corporation Journal of Applied Mathematics 2003:1(2003)55–64
2000 Mathematics Subject Classification: 17B10, 17B81, 15A90, 35Q40, 81V80 URL:http://dx.doi.org/10.1155/S1110757X03202047
[K3, K1] =rK1,[K3, K2] =−rK2,[K3, K4] =0,[K4, K1] =−tK1,[K4, K2] = tK2, subject to the physical conditions,K3 andK4 are real diagonal op- erators representing energy,K2=K†1, and the HamiltonianH=ω1K3+ (ω1+ω2)K4+λ(t)(K1e−iφ+K2eiφ)is a Hermitian operator. The Lie alge- braLsr,tis introduced as a generalization of the Tavis-Cummings model namely,L12,1 in [1]. Obviously, the subalgebra of Lsr,t generated byK1, K2, andK3in respective withK+,K−, andK0 is a generalization ofLsr, when dropping the physical condition(K++K−)must be real. That con- dition forced the representation matrices ofK+andK−to be real,[2,3,4].
Faithful matrix representations of least degree are discussed forLsr,tfor appropriate values ofr, s, t∈R.
Unless otherwise stated,Imis the identity matrix of degreem,Ois the zero matrix of appropriate size,N={1,2, . . . , n} andA= [aij],B= [bij], C= [δijcij], andD= [δijdij]aren×nreal matrices, where the matrices X=A+iB,Y =AT−iBT,C, and Dare representation matrices forK1, K2,K3, andK4, respectively;i=√
−1. All representations forLsr,tunder consideration are supposed to satisfy the above-mentioned physical re- quirements.
Lemma1.1. The Lie algebraLsr,tcan be defined by K1, K2
=sK3, K3, K1
=rK1, K4, K1
=−tK1, (1.1)
whereK3andK4are real diagonal operators andK2=K†1.
Proof. Indeed −rK2=−(rK1)†=−[K3, K1]†= [K3, K2]and similarly, for the relation[K4, K2] =tK2. SinceK3andK4are diagonal, they commute.
The Hermiticity of the Hamiltonian follows sinceω1, ω2, λ(t)∈R. As a necessity ofLemma 1.1we have the following lemma.
Lemma1.2. The matricesA,B,C, andDsatisfy the following:
(i) [A, BT]is a symmetric matrix, (ii) [A, AT] + [B, BT] =sC, (iii) [C, A] =rA,[C, B] =rB, (iv) [D, A] =−tA,[D, B] =−tB.
Lemma1.3. LetL,M, andKben×nmatrices such that[L, M] =aK,a=0, thentrace(K) =0.
Lemma1.4. Letp, q∈N,andσ= (pq)be a transposition. The representation obtained by applyingσto the rows as well as to the columns ofX,Y,C, andD is a conjugate representation forLsr,tand satisfies the physical requirements.
Proof. LetPbe the elementary matrix obtained by applyingσto the rows ofIn. SinceP=P−1=PT=P†, then the proof of the lemma follows.
Since[C, X] =rX, then for alli, j∈Nwe have, aij
cii−cjj−r
=0, bij
cii−cjj−r
=0. (1.2)
Similarly, fromLemma 1.2(iv), aij
dii−djj+t
=0, bij
dii−djj+t
=0. (1.3)
Ifxij=0, then from(1.2)and(1.3)
cii−cjj =r, djj−dii=t. (1.4)
Since[X, Y] =sC, then for eachi∈Nwe have,
scii=n
l=1
xil2−xli2
=n
l=1
a2il−a2li+b2il−bli2
. (1.5)
Lemma1.5. If t2+r2=0, then (1)xii=0, for alli∈N,
(2)ifxij=0thenxji=0, for alli, j∈N.
Proof. Ifr=0, then from(1.2)we have, for eachi∈N, thatxii=0. Also, ifxij=0, thencjj−cii−r=−2r, thusxji=0. Similarly, whent=0.
Lemma1.6. Ifs=0, then (1)trace(C) =0,
(2)ifxij=0then, fori, j∈N
r=1 s
n l=1
xil2−xli2−xjl2+xlj2
. (1.6)
Proof. Since[X, Y] =sCthen fromLemma 1.3, trace(C) =0. The proof of
(2), follows from(1.4)and(1.5).
We build the representation matrices starting withC.
Remark 1.7. UsingLemma 1.4,Ccan be rearranged intokdiagonal blo- cks, theith diagonal block consists of thekiscalar matrices,{ciImi,0,(ci− r)Imi,1, . . . ,[ci−r(ki−1)]Imi,(ki−1)}, wheremi,j is the repetitions of(ci−rj)
in the diagonal ofC; fori=1,2, . . . , kandj=0,1, . . . , ki−1. Thus, C=diag
c1Im1,0, c1−r
Im1,1, . . . , c1−r
k1−1
Im1,(k1−1), . . . , ciImi,0,
ci−r
Imi,1, . . . , ci−r
ki−1
Imi,(ki−1), . . . , ckImk,0,
ck−r
Imk,1, . . . , ck−r
kk−1 Imk,(kk−1)
,
(1.7)
where
ci=cj, wheneveri=j, fori, j=1,2, . . . , k, (1.8) ci−rj
−ci+1=r, forj=0, . . . , ki−1; i=1,2, . . . , k−1. (1.9) The ith diagonal block of C is called the ci-block and ki is its length.
Any diagonal entrycofCsuch thatc=ci−rl, forl≥0 then 0≤l≤ki− 1 for somei=1, . . . , k, that is,c belongs to theci-block. If ci−l1r=cj− l2r, 0≤l1 ≤ki−1, 0≤l2≤kj−1, then ci and cj are in the same block, violating(1.9).
We use the notations given inRemark 1.7.
2. Faithful representations forLsr,twherers=0
Lemma2.1. The matricesAandBcan be partitioned into submatrices of the same size corresponding to those ofC.The nonzero submatrices ofAandBare all off-diagonal submatrices.
Proof. From(1.2), the diagonal submatrices ofAandBare square zero submatrices of ordersm1,0, . . . , mk,(kk−1), in respective to those ofC. Let cii, cjj, and cll; i, j, l∈N, be from different diagonal submatrices of C, and suppose thataij =0 andail=0, then from(1.2),cll=cjj contradict- ing(1.8). Similarly, ifajiandaliare from different submatrices inAthey cannot be both nonzero. In view of(1.2), only the off-diagonal subma- trices ofAmay be nonzero. Thus we have,A= [Aij]whereAij=O, for
j=i+1. And similarly forB.
Lemma2.2. Fork >1, if ki=1, for somei=1,2, . . . , k, thenLsr,thas a repre- sentation of degreen−mi,0. Moreover, if the entries in theith row and theith column of Xare all zeros, thenLsr,thas a representation of degreen−1.
Proof. We use Lemma 1.4 so that the ci-block becomes the first block of the main diagonal ofC. Since for allj∈N, 1≤i≤m1,0,|cii−cjj| =r, otherwise ki>1, then from (1.2) the representation is fully reducible since, A=0 0
0A
, B=0 0
0B
, C=C 1 0 0 C2
, andD=D 1 0 0 D2
. The matrices
X=A+iB,Y=X†,C2, andD2 are all of degreen−mi,0and satisfy the lemma. Similar argument holds when the entries in theith row and the
ith column ofXare all zeros.
So, it can be assumed that ifk >1 thenki>1;i=1, . . . , k. And forX= O, if the entries of theith row of Xare all zeros, then those of theith column are not all zeros, and vice versa, in such cases, we get from(1.5) thatscii=0.
Theorem2.3. Ifrs <0, thenX=Y =C=O.
Proof. Ifk=1 andk1=1, then from(1.2)X=Y =O. IfX=O, then from (1.5) C=O. Suppose that X=O, there are only two cases to consider namely, the case where k=1 and k1>1, and the case wherek >1. In both casesk1>1, from Lemma 2.1the first m1,0 columns of Xare zero columns, and fromLemma 2.2there must be anx1,j =0 for somem1,0<
j≤(m1,0+m1,1). Thus from(1.5),
sc11=sc1=n
l=1
x1l2−0
>0. (2.1)
Letα=m1,0+m1,1+···+m1,(k1−2). Ifk >1, we get from(1.9),[c1−r(k1− 1)]−c2=r, thus from(1.2), the rowsα+1, α+2, . . . , α+m1,(k1−1) are zero rows of X. If k=1 and k1>1, we get from Lemma 2.1 that the men- tioned rows are zero rows ofX, being the last rows ofX. In both cases, fromLemma 2.2there must be anxi,α+1=0 for some[α−m1,(k1−2)]< i≤α.
From(1.5),
scα+1,α+1=s c1−r
k1−1
=n
l=1
0−xl,α+12
<0. (2.2)
Ifs >0, thenc1>0 by(2.1), sincer <0, then[c1−r(k1−1)]>0, violat- ing(2.2). Similarly, ifs <0, we get from(2.1),[c1−r(k1−1)]<0, violat-
ing(2.2).
We conclude this section by introducing the 2×2 representation ma- tricesX,Y,C, andDofK1,K2,K3, andK4, respectively, forrs >0,t∈R
X= 0 a±i
rs/2−a2
0 0
, Y = 0 0
a∓i
rs/2−a2 0
,
C=
r/2 0 0 −r/2
, D=
b 0 0 b+t
,
(2.3)
for anya, b∈Rsuch that|a| ≤
rs/2 and for the linear independency of CandD, takeb=−t/2. These representations are faithful. The 2×2 representation matricesX,Y,C, andDgeneralize those given in[1].
Clearly, the vector space spanned byX,Y, andCissl(2,C), as a vector space. The representation matrices ofLsr, in[2], are for the special cases, a2=rs/2.
3. Faithful representations forLsr,twhererst=0
The case wherers=0 andt=0 was considered in the previous section.
So, ifs=0 we only need to consider the case wherer=0 andtis any real number.
3.1. Fors=0,r=0, andt∈R
Sincer=0 then anyci-block of the matrixC has lengthki=1. So, we haveC=diag(c1Im1, . . . , ckImk)whereci=cjwheneveri=j;i, j=1, . . . , k.
Remark 3.1. IfXcommutes withY =X†, thenXis a normal matrix, and there exists a unitary matrixUsuch thatX=U†ZUfor some complex diagonal matrixZ. IfUcommutes withCandD, then the diagonal ma- tricesZ, ¯Z,C, andDare representation matrices forK1,K2,K3, andK4, respectively, and satisfy the physical requirements. We takeU=Inwhen Xis diagonal.
Lemma3.2. If C=diag(c1Im1, . . . , ckImk)for differentci’s, then the represen- tation is fully reducible into representations of degreesm1, . . . , mk.
Proof. The matrixD is diagonal and from (1.2),xij =xji=yij=yji=0,
whenevercii=cjj;i, j∈N.
Lemma3.3. LetK= [Kij]be a partitioned matrix which is normal whose di- agonal blocks areksquare matrices. IfKij=Owheneverj=i+1(orj=i−1);
i, j=1, . . . , k. ThenK=O.
Proof. LetK= [kij]be ann×nmatrix, then for eachi∈N, n
l=1
kil2=n
l=1
kli2. (3.1)
Let the diagonal blocks ofKbe of degreesi1, . . . , ik, respectively. IfKij=O wheneverj=i+1;i, j=1, . . . , k, then the firsti1rows ofKare zeros, thus from(3.1)the firsti1columns ofKare zeros. Continuing like that in less
thanksteps, it can be shown thatK=O. Hence the proof of the lemma
follows.
Theorem3.4. The matrixC=O, in any representation ofLs0,t. Ifst=0, then X=Y =O.
Proof. SupposeC=O, we useLemma 1.4 so thatc1=0, from(1.5)and Lemma 3.2,m1sc1=m1
i=1scii=m1
i=1
m1
l=1(|xil|2− |xli|2) =0, butm1sc1=0.
ThenC=O. Thus fromLemma 1.1,Xis a normal matrix. Ift=0, we use Lemma 1.4, so that
D=diag
d1Im1,0, d1+t
Im1,1, . . . , d1+t
k1−1 Im1,(k
1−1), . . . , diImi,0, di+t
Imi,1, . . . , di+t
ki−1 Imi,(k
i−1), . . . , dkImk,0, dk+t
Imk,1, . . . , dk+t
kk−1 Im
k,(k k −1)
,
(3.2)
wheremi,jis the repetitions of(di+tj)in the diagonal ofD; fori=1, . . . , k andj=0, . . . , ki−1 such that
di=dj, wheneveri=j, fori, j=1,2, . . . , k, di+1−
di+tj
=t, forj=0, . . . , ki−1; i=1,2, . . . , k−1. (3.3)
From(1.3),Xcan be partitioned into submatrices of the same sizes cor- responding to those ofD, whose nonzero submatrices are off-diagonal
submatrices. Then byLemma 3.3X=Y =O.
Ift=0 then fromLemma 1.1, the generators commute and such a case can be considered as a special case ofL00,0ofSection 3.3, withC=O.
3.2. Fors=0andr2+t2=0
From (1.5) as s=0, then(3.1) holds. If the ith row (or column) of X consists entirely of zeros, theith column(or row)also, consists entirely of zeros and both can be omitted by the following lemma whose proof is analogous to that ofLemma 2.2. So, ifX=O, it can be considered thatX has no zero row or zero column.
Lemma3.5. If Xhasmzero rows (or columns), where0≤m < n, thenLsr,thas a representation of degreen−m.
Theorem3.6. Ifs=0andr2+t2=0,Lsr,thas no faithful representations. In any representation,X=Y=O.
Proof. Ifr=0, arrangeCas inRemark 1.7otherwise, letDas in the proof ofTheorem 3.4. In view ofLemma 1.5,Xcan be partitioned into subma- trices of the same sizes corresponding to those of C when r=0 or to those ofD otherwise. The nonzero submatrices of X are all off diago- nal submatrices. Ass=0 thenXis normal and fromLemma 3.3, we get
X=Y =O.
3.3. Fors=r=t=0
Although physically is not applicable, but for the sake of completeness, we consider the case whenK1, K2, K3, andK4are commutant operators.
Theorem 3.7. The representations of L00,0 are conjugate to representations whereK1,K2,K3, andK4are represented by diagonal matrices.
Proof. LetX=U†ZUfor a unitary matrixUand a complex diagonal ma- trixZ. We claim thatUcommutes withCandD, then the theorem holds by usingRemark 3.1. We induce onn, the degree of the representation and prove the cases whenXis not diagonal.
Forn=2: ifXis not diagonal then from(1.4), bothCandDare scalar matrices and both commute withU.
Forn=3: if the diagonal elements ofC(orD)are all different, thenX must be diagonal. IfXhas two nonzero elementsxijandxlm, from(1.4), both are nondiagonal elements wherexlmis not thexji, thenCandDare scalar matrices and both commute withU. Otherwise, we use Lemma 1.4, so that X=XO
O g
, thus from (1.2) and (1.3) C=cI2O
O a
and D= dI2O
O b
, for some a, b, c, d∈R;g∈C, whereXis not a diagonal matrix.
That requiresXto be a normal matrix. So, there exists a unitary matrix Usuch thatX=U†MU, for some complex diagonal matrixM. Obvi- ously,Ucommutes withcI2anddI2. LetU=UO
O 1
, andZ=diag(M, g) thenUcommutes withCandD.
Assume that the theorem is true forn < m.
Forn=m: if bothCandDare scalar matrices, thenUcommutes with C and D. If either C or D is not a scalar matrix, C say, then we use Lemma 1.4to rearrangeCso thatC=diag(c1Im1, . . . , ckImk)for different cis, from(1.2)X=diag(X1, . . . , Xk)whereXi is a square matrix of order mi< m. Also,Dcan be considered asD=diag(D1, . . . , Dk)whereDiis a diagonal matrix of degreemi. Hence, the representation is fully reducible into representations of degreesmi,i=1, . . . , k. SinceXis normal thenXi
is normal fori=1, . . . , k. Thus there exists a unitary matrixUisuch that
Xi=U†iZiUifor some complex diagonal matrixZi,i=1, . . . , k. From the inductionUicommutes withciImi andDi. LetU=diag(U1, . . . , Uk)and Z=diag(Z1, . . . , Zk), thenUcommutes withCandD.
Theorem3.8. The Lie algebraL00,0 has faithful representations of degree4as the least degree.
Proof. Any linearly independent diagonal matrices Z, ¯Z,C, and D, of degree 4, withCandDare real, are representation matrices forK1,K2, K3, andK4, respectively, of a faithful representation.
We conclude the paper by mentioning the cases whereLsr,thas faithful matrix representations satisfying the physical requirements.
Summary3.9. It is assumed that all representations of Lsr,t must satisfy the physical requirements.
(1)For rs >0,t∈R,Lsr,t has faithful representations of degree 2as the least degree.
(2)For r =s=t=0, L00,0 has faithful representation of degree 4as the least degree where the representation matrices are linearly independent diagonal matrices, withCandDare real matrices.
Acknowledgments
The author is grateful to Prof. S. S. Hassan, Ain Shams University for suggesting the problem and fruitful discussions and to Prof. S. Singh, King Saud University for reading the manuscript. The support of Kuwait University is also appreciated.
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L. A. M. Hanna: Department of Mathematics and Computer Science, Faculty of Science, Kuwait University, P.O. Box 5969, Safat 13060, Kuwait
E-mail address:[email protected]
