An Approximate Solution of the Jaynes Cummings Model with Dissipation

Transcription

1 An Approximate Solution of the Jaynes Cummings Model with Dissipation arxiv:03.039v [quant-ph] Mar 0 Kazuyuki FUJII and Tatsuo SUZUKI Department of Mathematical Sciences Yokohama City University Yokohama, Japan Center for Educational Assistance Shibaura Institute of Technology Saitama, Japan Abstract In this paper we treat the Jaynes Cummings model with dissipation and give an approximate solution to the master equation for the density operator under the general setting by making use of the Zassenhaus expansion. In this paper we treat the Jaynes Cummings model([]) with dissipation(or the quantum damped Jaynes Cummings model in our terminology) and study the structure of general solution from a mathematical point of view in order to apply it to Quantum Computation address : fujii@yokohama-cu.ac.jp address : suzukita@aoni.waseda.jp ; i070@sic.shibaura-it.ac.jp

2 or Quantum Control, which are our final target. As a general introduction to these topics see for example [] and [3]. We believe that the model will become a good starting point to study more sophisticated models (with dissipation) in a near future, see for example Concluding Remarks in the paper. et us start with the following phenomenological master equation for the density operator of the atom cavity system in [4] : t ρ = i[h JC,ρ]+µ aρa (a aρ+ρa a) } +ν a ρa } (aa ρ+ρaa ) () where H JC is the well known Jaynes-Cummings Hamiltonian (see []) given by H JC = ω 0 σ 3 +ω 0 a a+ω ( σ + a+σ a ) () ω 0 = +ω 0 Ωa Ωa ω 0 +ω 0 with σ + = 0 0 0, σ = 0 0 0, σ 3 = 0 0, = 0 0, and a and a are the annihilation and creation operators of an electro magnetic field mode in a cavity, a a is the number operator, and µ and ν (µ > ν 0) are some constants depending on it (for example, a damping rate of the cavity mode). ote that the density operator ρ is in M(;C) M(F) = M(;M(F)), namely ρ = ρ 00 ρ 0 M(;M(F)) (3) ρ 0 ρ where M(F) is the set of all operators on the Fock space F defined by n=0 and is the identity operator. F Vect C 0,,, 3, } } = c n n c n < ; n = (a ) n 0 n! n=0

3 We would like to solve () explicitly. However, it is very hard to solve at the present time, so we must satisfy by giving some approximate solution to it. First of all we write down the equation () in a component-wise manner. For that we set for simplicity. Then it is easy to see H JC = ω 0 +ω 0 Ωa Ωa ω 0 +ω 0 A B (4) C D ρ 00 = i(aρ 00 +Bρ 0 ρ 00 A ρ 0 C)+ µ aρ 00 a } (a aρ 00 +ρ 00 a a) +ν ρ 0 = i(aρ 0 +Bρ ρ 00 B ρ 0 D)+ µ aρ 0 a } (a aρ 0 +ρ 0 a a) +ν ρ 0 = i(cρ 00 +Dρ 0 ρ 0 A ρ C)+ µ aρ 0 a } (a aρ 0 +ρ 0 a a) +ν a ρ 00 a } (aa ρ 00 +ρ 00 aa ), a ρ 0 a } (aa ρ 0 +ρ 0 aa ), a ρ 0 a } (aa ρ 0 +ρ 0 aa ), ρ = i(cρ 0 +Dρ ρ 0 B ρ D)+ µ aρ a } (a aρ +ρ a a) +ν a ρ a } (aa ρ +ρ aa ) (5) where ρ ij = ( / t)ρ ij as usual. Here let us remind some technique used in [5] and [6], which is very useful in some case. For a matrix X = (x ij ) M(F) X = x 00 x 0 x 0 x 0 x x x 0 x x we correspond to the vector X F dim CF as X = (x ij ) X = (x 00,x 0,x 0, ;x 0,x,x, ;x 0,x,x, ; ) T (6) 3

4 where T means the transpose. Then the following formula EXF = (E F T ) X (7) holds for E,F,X M(F). This and equations (5) give ˆρ 00 = i ( A A T) ˆρ 00 C Tˆρ } 0 +B ˆρ 0 + [ µ a (a ) T } (a a + a a) +ν a a T }] (aa + aa ) ˆρ 00, ˆρ 0 = i B Tˆρ 00 + ( A D T) } ˆρ 0 +B ˆρ + [ µ a (a ) T } (a a + a a) +ν a a T }] (aa + aa ) ˆρ 0, ˆρ 0 = i C ˆρ 00 + ( D A T) ˆρ 0 C Tˆρ } + [ µ a (a ) T } (a a + a a) +ν a a T }] (aa + aa ) ˆρ 0, ˆρ = i C ˆρ 0 B Tˆρ 0 + ( D D T) } ˆρ + [ µ a (a ) T } (a a + a a) +ν a a T }] (aa + aa ) ˆρ (8) because and = a a are diagonal ( T =, T = ). From (3) we set ˆρ = ˆρ 00 ˆρ 0 ˆρ 0 = ρ = ρ 00 ρ 0 ρ 0 ρ, (9) ˆρ then we obtain the following canonical form 4

5 = i tˆρ + A A T C T B 0 B T A D T 0 B ˆρ C 0 D A T C T 0 C B T D D T ˆρ (0) with = µ a (a ) T } (a a + a a) +ν a a T } (aa + aa ), or more explicitly tˆρ = ω 0 ( ) Ω (a ) T Ωa 0 Ω a i T ω 0 +ω 0 ( ) 0 Ωa Ωa 0 ω 0 +ω 0 ( ) Ω (a ) T ˆρ 0 Ωa Ω a T ω 0 ( ) + ˆρ () from (4) : A = ω 0 +ω 0, B = Ωa, C = Ωa, D = ω 0 +ω 0. Here a simplified notation ω 0 in place of ω 0 has been used. 5

6 ow we set ω 0 ( ) Ω (a ) T Ωa 0 Ω a X = T ω 0 +ω 0 ( ) 0 Ωa Ωa 0 ω 0 +ω 0 ( ) Ω (a ) T, 0 Ωa Ω a T ω 0 ( ) Y = ( = µ a (a ) T } (a a + a a) +ν a a T }) (aa + aa ) () (3) for simplicity. ext, in order to rewrite X and Y in terms of ie algebraic notations used in [5] we set K + = a a T, K = a (a ) T, K 3 = ( + + ), K 0 =. (4) Then it is easy to see (K + ) = K, (K 3 ) = K 3, (K 0 ) = K 0, [K 3,K + ] = K +, [K 3,K ] = K, [K +,K ] = K 3, [K 0,K + ] = [K 0,K ] = [K 0,K 3 ] = 0 (5) and [K 0,a ]+a = 0, [K 0, a T ]+ a T = 0, [K 0,a ] a = 0, [K 0, (a ) T ] (a ) T = 0. (6) amely, K +,K,K 3 } are generators of the ie algebra su(,), see for example [7] as a general introduction. Under the above notations X and Y in () and (3) can be rewritten as 6

7 and ω 0 K 0 Ω (a ) T Ωa 0 Ω a T ω 0 +ω 0 K 0 0 Ωa X = Ωa 0 ω 0 +ω 0 K 0 Ω (a ) T 0 Ωa Ω a T ω 0 K 0 Y = µ ν ( = νk + +µk (µ+ν)k 3 + µ ν ) (7) +(νk + +µk (µ+ν)k 3 ) 4 (for simplicity) (8) where in the process the relation aa = a a+ = + has been used. As a result we obtain the final form = ( ix +Y)ˆρ (9) tˆρ with X in (7) and Y in (8). This is our main result and we believe that the equation is clear cut enough (compare it with ()). Since the general solution is given by ˆρ(t) = e t( ix+y )ˆρ(0) (0) in a formal way we must calculate the term e t( ix+y ), which is in general very hard. For that the following Zassenhaus formula is convenient. Zassenhaus Formula We have an expansion e t(a+b) = e t3 6 [[A,B],B]+[[A,B],A]} e t [A,B] e tb e ta. () The formula is a bit different from that of [8]. In this paper we use the approximation e t(a+b) e t [A,B] e tb e ta 7

8 with A = ix and B = Y, so that ˆρ(t) is approximated as ˆρ(t) e it [X,Y] e ty e itxˆρ(0). () et us calculate each term explicitly : [I] First, we calculate e itx. However, the calculation is more or less well known. We decompose X into two parts ω 0 K 0 Ω (a ) T 0 0 Ω a T ω 0 +ω 0 K X =, 0 0 ω 0 +ω 0 K 0 Ω (a ) T 0 0 Ω a T ω 0 K Ωa Ωa X =. Ωa (3) 0 Ωa 0 0 Then it is not difficult to check [X,X ] = 0 (4) owing to the relations (6). If we set S = (5) then SX S = SX S = Ω 0 a 0 0 a a 0 0 a 0. (6) 8

9 Moreover, if we decompose like ω 0K 0 Ω (a ) T = ω 0K Ω a T ω 0 +ω 0 K 0 0 ω 0 +ω 0 K 0 then 0K 0 0, 0 ω 0 +ω 0 K 0 0 Ω (a ) T Ω a T 0 0 Ω (a ) T Ω a T 0 = 0 owing to the relations (6). Therefore we have a decomposition ω 0 K ω 0 +ω 0 K X = 0 0 ω 0 +ω 0 K ω 0 K 0 0 (a ) T 0 0 a T Ω (a ) T (7) 0 0 a T 0 As a result we have a decomposition consisting of three commutative operators ω 0 K ω 0 +ω 0 K X = 0 0 ω 0 +ω 0 K ω 0 K 0 0 (a ) T 0 0 a T Ω (a ) T 0 0 a T 0 0 a 0 0 a Ω S S a (8) 0 0 a 0 9

10 ow, by making use of the well known formula exp iα 0 A = A 0 cos(α AA ) i sin(α AA AA )A i A A sin(α A A)A cos(α A A) where α is some parameter, we can calculate e itx easily. The result is () () (3) (4) e itx () () (3) (4) = (3) (3) (33) (34) (4) (4) (43) (44) (9) (30) where for α = Ωt () = e itω 0K 0 cos(α aa )cos(α aa ) = e itω 0 cos(α +) e itω 0 cos(α +), () = i e itω 0K 0 aa sin(α aa )( (a ) T )cos(α aa ) = i e itω 0 cos(α +) e itω 0 + sin(α +)(a ) T, (3) = i e itω 0K 0 cos(α aa ) aa sin(α aa )(a ) = i e itω 0 + sin(α +)a e itω 0 cos(α +), (4) = e itω 0K 0 sin(α aa )( (a ) T ) aa aa sin(α aa )(a ) = e itω 0 + sin(α +)a e itω 0 + sin(α +)(a ) T ; and 0

11 () = ie itω 0 e itω 0K 0 a a sin(α a a)( a T )cos(α aa ) = ie itω 0 e itω 0 cos(α +) e itω 0 sin(α )a T, () = e itω 0 e itω 0K 0 cos(α a a)cos(α aa ) = e itω 0 e itω 0 cos(α +) e itω 0 cos(α ), (3) = e itω 0 e itω 0K 0 a a sin(α a a)( a T ) aa sin(α aa )(a ) = e itω 0 e itω 0 + sin(α +)a e itω 0 sin(α )a T, (4) = ie itω 0 e itω 0K 0 cos(α a a) aa sin(α aa )(a ) = ie itω 0 e itω 0 + sin(α +)a e itω 0 cos(α ); and (3) = ie itω 0 e itω 0K 0 cos(α aa ) a a sin(α a a )(a ) = ie itω 0 e itω 0 sin(α )a e itω 0 cos(α +), (3) = e itω 0 e itω 0K 0 sin(α aa )( (a ) T ) aa a a sin(α a a )(a ) = e itω 0 e itω 0 sin(α )a e itω 0 + sin(α +)(a ) T, (33) = e itω 0 e itω 0K 0 cos(α aa )cos(α a a ) = e itω 0 e itω 0 cos(α ) e itω 0 cos(α +), (34) = ie itω 0 e itω 0K 0 aa sin(α aa )( (a ) T )cos(α a a ) = ie itω 0 e itω 0 cos(α ) e itω 0 + sin(α +)(a ) T ; and

12 (4) = e itω 0K 0 a a sin(α a a)( a T ) a a sin(α a a )(a ) = e itω 0 sin(α )a e itω 0 sin(α )a T, (4) = ie itω 0K 0 cos(α a a) a a sin(α a a )(a ) = i e itω 0 sin(α )a e itω 0 cos(α ), (43) = ie itω 0K 0 a a sin(α a a)( a T )cos(α a a ) = i e itω 0 cos(α ) e itω 0 sin(α )a T, (44) = e itω 0K 0 cos(α a a)cos(α a a ) = e itω 0 cos(α ) e itω 0 cos(α ). [II] Second, we must calculate e ty. To be glad this task has been done, see [5] and [6]. amely, from (8) we have e t e ty e t = e t e t ; = µ ν +νk + +µk (µ+ν)k 3 (3) and e t = e µ ν t e tνk ++µk (µ+ν)k 3 } = e µ ν t e G(t)K + e log(f(t))k 3 e E(t)K (3) with E(t) = µ sinh( µ ν t) µ ν cosh ( µ ν t) + µ+ν ( ) µ ν F(t) = cosh t G(t) = µ ν sinh( µ ν + µ+ν µ ν sinh ν sinh( µ ν t) µ ν cosh ( µ ν t) + µ+ν sinh( µ ν µ ν t), ( µ ν ) t, t). (33)

13 This is a kind of disentangling formula, see for example [7] as a general introduction. [III] Third, we must calculate e it [X,Y]. In this stage some interaction appears. et us calculate [X, Y] exactly. Some calculation gives [ a T,νK + +µk (µ+ν)k 3 ] = µa + µ+ν a T, [ (a ) T,νK + +µk (µ+ν)k 3 ] = νa µ+ν (a ) T, [a,νk + +µk (µ+ν)k 3 ] = ν a T µ+ν a, [a,νk + +µk (µ+ν)k 3 ] = µ (a ) T + µ+ν a, [K 0,νK + +µk (µ+ν)k 3 ] = 0 (34) and for simplicity we set A = νa + µ+ν (a ) T, B = µa µ+ν a T, C = ν a T µ+ν a, D = µ (a ) T + µ+ν a. (35) ote that B A and D C because of µ > ν. It is easy to see [A,C] = [A,D] = 0, [B,C] = [B,D] = 0, ( ) µ ν [A,B] = [C,D] =. (36) From the decomposition (8) we have a decomposition consisting of two commutative operators [X,Y] = Ω 0 A 0 0 B A 0 0 B 0 +ΩS 0 C 0 0 D C 0 0 D 0 S (37) with S in (5). 3

14 ow, by making use of the formula exp iβ 0 P Q 0 = cos(β PQ) i QP sin(β QP)Q i PQ sin(β PQ)P cos(β QP) (38) for some parameter β, we can calculate the term e it [X,Y] easily. The result is () () (3) (4) e it [X,Y] () () (3) (4) = (3) (3) (33) (34) (4) (4) (43) (44) (39) where for β = Ω t () = cos(β AB)cos(β CD), () = i sin(β AB)Acos(β CD), AB (3) = icos(β AB) sin(β CD)C, CD (4) = AB sin(β AB)A CD sin(β CD)C ; () = i sin(β BA)Bcos(β CD), BA () = cos(β BA)cos(β CD), (3) = BA sin(β BA)B CD sin(β CD)C, (4) = icos(β BA) sin(β CD)C ; CD (3) = icos(β AB) sin(β DC)D, DC (3) = AB sin(β AB)A DC sin(β DC)D, (33) = cos(β AB)cos(β DC), (34) = i sin(β AB)Acos(β DC) ; AB 4

15 (4) = BA sin(β BA)B DC sin(β DC)D, (4) = icos(β BA) sin(β DC)D, DC (43) = i sin(β BA)Bcos(β DC), BA (44) = cos(β BA)cos(β DC) where AB = µνa a + µ+ν BA = µνaa +µ µ+ν CD = µν a a+ µ+ν DC = µν aa +µ µ+ν These are complicated enough. νa a T +µ µ+ν a (a ) T a (a ) T + µ+ν νa a T νa a T +µ µ+ν a (a ) T a (a ) T + µ+ν νa a T ( ) µ+ν aa, ( ) µ+ν a a, ( ) µ+ν aa, ( ) µ+ν a a. Anyway, we have completed the task although it is very complicated. In last, we shall restore the result to original form. If we set ˆ ρ(t) = e ty e itxˆρ(0) e ty ˆ ρ (t), (40) then ρ (t) is from (30) given by where ρ (t) = () () () () (4) 5

16 () = e itω 0 cos(α +) ρ 00 e itω 0 cos(α +) +ie itω 0 cos(α +) ρ 0 a e itω 0 + sin(α +) ie itω 0 + sin(α +)a ρ 0 e itω 0 cos(α +) +e itω 0 + sin(α +)a ρ a e itω 0 + sin(α +), () = ie itω 0 e itω 0 cos(α +) ρ 00 ae itω 0 sin(α ) +e itω 0 e itω 0 cos(α +) ρ 0 e itω 0 cos(α ) +e itω 0 e itω 0 + sin(α +)a ρ 0 ae itω 0 sin(α ) ie itω 0 e itω 0 + sin(α +)a ρ e itω 0 cos(α ), () = ie itω 0 e itω 0 sin(α )a ρ 00 e itω 0 cos(α +) +e itω 0 e itω 0 sin(α )a ρ 0 a e itω 0 + sin(α +) +e itω 0 e itω 0 cos(α ) ρ 0 e itω 0 cos(α +) +ie itω 0 e itω 0 cos(α ) ρ a e itω 0 + sin(α +), () = e itω 0 sin(α )a ρ 00 ae itω 0 sin(α ) ie itω 0 sin(α )a ρ 0 e itω 0 cos(α ) +ie itω 0 cos(α ) ρ 0 ae itω 0 sin(α ) +e itω 0 cos(α ) ρ e itω 0 cos(α ), and ρ(t) is by ρ(t) = eµ ν t F(t) G(t) n (a ) n E(t) m [exp( log(f(t))}) a m ρ (t)(a ) } m n! m! n=0 m=0 exp( log(f(t))})]a n (4) in terms of E(t),F(t),G(t) in (33) and = a a. See [5] and [6]. This form is relatively clear because of no interaction. 6

17 Our (approximate) solution is ˆρ(t) = e it [X,Y] e ty e itx = e [X,Y]ˆ ρ(t) it from (40). We would like to express ρ(t) like (4). From (39) we can expand each term in terms of the Taylor expansion of cos(x) and sin(x). However, the form is very ugly in the Dirac s sense, so we leave such a task to readers. Concluding Remarks In this paper we treated the master equation for the density operator based on the Jaynes Cummings Hamiltonian with dissipation and constructed the approximate solution up to O(t 3 ) under the general setting. Our construction is quite general because ˆρ(0) is any initial state. However, it may be still inconvenient to apply to realistic problems coming from the atom cavity system. Further work will be required. Moreover, it may be possible to treat a generalized master equation for the density operator given by t ρ = i[h JC,ρ]+µ aρa } (a aρ+ρa a) +ν a ρa } (aa ρ+ρaa ) + κ aρa } (a ρ+ρa ) + κ a ρa } ((a ) ρ+ρ(a ) ) (43) with the positivity condition µν κ. However, we don t treat this general case in the paper, see [9], [0] and []. References [] E. T. Jaynes and F. W. Cummings : Comparison of Quantum and Semiclassical Radiation Theories with Applications to the Beam Maser, Proc. IEEE, 5 (963), 89. [] H. -P. Breuer and F. Petruccione : The theory of open quantum systems, Oxford University Press, ew York, 00. 7

18 [3] W. P. Schleich : Quantum Optics in Phase Space, WIEY VCH, Berlin, 00. [4] M. Scala, B. Militello, A. Messina, S. Maniscalco, J. Piilo and K.-A. Suominen : Cavity losses for the dissipative Jaynes-Cummings Hamiltonian beyond Rotating Wave Approximation, J. Phys. A: Math. Theor. 40 (007), 457, arxiv: [quant-ph]. [5] R. Endo, K. Fujii and T. Suzuki : General Solution of the Quantum Damped Harmonic Oscillator, Int. J. Geom. Meth. Mod. Phys, 5 (008), 653, arxiv : [quant-ph]. [6] K. Fujii and T. Suzuki : General Solution of the Quantum Damped Harmonic Oscillator II : Some Examples, Int. J. Geom. Meth. Mod. Phys, 6 (009), 5, arxiv : [quant-ph]. [7] K. Fujii : Introduction to Coherent States and Quantum Information Theory, quant-ph/0090. [8] C. Zachos : Crib otes on Campbell-Baker-Hausdorff expansions, unpublished, 999, see [9] K. Fujii : Algebraic Structure of a Master Equation with Generalized indblad Form, Int. J. Geom. Meth. Mod. Phys, 5 (008), 033, arxiv : [quant-ph]. [0] K. Fujii : A Master Equation with Generalized indblad Form and a Unitary Transformation by the Squeezing Operator, arxiv : [quant-ph]. [] K. Fujii : in progress. 8