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Detailed Derivation of the General Master Equation in Quantum Optics

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Detailed Derivation of the General Master Equation in Quantum Optics DETAILED DERIVATION OF THE GENERAL MASTER EQUATION IN QUANTUM OPTICS TIMOTHY JONES Abstract. An unoriginal but more detailed derivation that typical is pre- sented of the master equation. 1. Assumptions We begin by outlining the assumptions which enable the canonical derivation of the master equation. Our general assumption is that our system is a harmonic † † oscillator (aa ) which interacts with a bath of oscillators ( i bibi ). Specifically: (1) We assume that the energy spectrum of the bathP of oscillators is spaced tightly enough relative to its overall domain, i.e. ωi ≪ (ωN − ω0) such that the approximation i → dωig(ωi), where g(ωi) is the density of states of the reservoir, is a physically acceptable mathematical assumption. (2) Related to the previousP assumption,R we assume that the bath is large enough and in a state of equilibrium such that any perturbations caused by the individual system on the bath is negligible. This is to say, the fu- ture state of the system-bath density operator is determined by its current state, and is not a function of the history of the bath (that is, we assume ρsb = ρs(t) ⊗ ρb(0)).This is the Markoffian assumption. (3) We assume the rotating wave approximation regime. The interaction of the bath and system will have terms such as ab,a†b,ab†,a†b†. The lowering- lowering and raising-raising coupling has a much slower varying contribu- tion to the state of the system, and so are excluded to give the interaction † † Hamiltonian i gi a bi + abi where gi is a real coupling constant. Obvi- ously, one must be certain the system one models can be simplified as such P in order to apply the general master equation. 2. Construction of the Hamiltonian and Density Operators We assume the system can be modeled with the Hamiltonian, † † (2.1) Hs = ~ωa a, [a,a ]=1 Separately we assume that there exists a bath which can be modeled with the Hamiltonian, ~ † † (2.2) Hb = ωibi bi, [bi,bk]= δik i X Finally, as stated before, the interaction of the bath and system is taken in the form of † † (2.3) gi a bi + abi , gi ∈ℜ i X 1 The total Hamiltonian is, H0 ~ † ~ † † † (2.4) H = ωa a + ωj bjbj + gi(a bi + abi ) j i z X}| { X HI We assume uncertainty in the preparation of| states,{z so we swi} tch to the density operator regime. We call the density operator corresponding to our bath coupled system ρSB (S: Single original oscillator; B: Environment) where the individual density operators can be retrieved via a trace, i.e. ρS = T rB(ρSB)= hB|ρSB |Bi Trace over environment states B X ρB = T rS(ρSB)= hS|ρSB|Si Trace over local states S X The dynamics of the system evolve as (Liouville equation), dρ (2.5) i~ SE = [H,ρ ] dt SE We commit a unitary transform to simplify this equation as follows (the so-called interaction picture). It can be shown that, α2 (2.6) exp(αA)B exp(−αA)= B + α[A, B]+ [A, [A, B]] + · · · 2! Furthermore, from the premises of Quantum Mechanics, it is legitimate to commit transforms of the type, (2.7) G′ =ΛGΛ†, Λ ∈ Unitary These so called similarity transformations preserve rank, determinant, trace, and eigenvalues. We let, i i (2.8) ρse = exp ~Hot ρSE exp −~Hot We take the time derivative and find, dρ i i i i (2.9) se = [H ,ρ ] − exp H t [H,ρ ]exp − H t dt ~ o se ~ ~ o SE ~ o From equation 2.6, since ΛGHΛ† =ΛGΛ†ΛHΛ† and [Ho, Ho]=0, i i exp ~Hot Ho exp −~Hot = Ho dρ i (2.10) se = − [H ,ρ ] dt ~ i se where Hi is the transform of HI and can be calculated as follows. We know that, [a†a,a†]= a†aa† − a†a†a = [a,a†]a† = a† [a†a,a]= a†aa − aa†a = [a†,a]a = −a † and whereas Ho = ~ωa a, equation 2.6 gives exp αa†a a† exp(−αaa†) = exp(α)a†, † † exp αa a a exp(−αaa ) = exp(−α)a, α ∈ℑ 2 † † Hi can be calculated in this fashion for both the aa and the multiple set bibi , where it is thus now trivial to find that, ~ † † (2.11) Hi(t)= gi a bj exp((i(ω − ωi)t)+ abi exp(−i(ω − ωi)t) i X If we now define G(t) ≡ gibi exp(i(ω − ωj )t) i X we can simplify to † † (2.12) Hi(t)= ~ G(t)a + G (t)a Since the bath is in equilibrium, we can construct ρe(0) easily. Let Zi = exp(−~ωi/kT ). Then the probability that one mode (i) of the field is excited with n photons is Zn −~ω −n~ω (2.13) P = i = (1 − Z )Zn = 1 − exp i exp i i,n Zn i i kT kT n i The density matrixP for this mode is then −~ω −n~ω (2.14) ρ = 1 − exp i exp i |nihn| e,i kT kT n X −~ω −b†b ~ω = 1 − exp i exp i i i |nihn| kT kT n ! X To simplify this further, we imagine a three states system in which we can represent b†b|ni = n|ni as, for example letting n = 2, 0 0 0 0 0 0 1 0 0 =2 0 0 0 2 1 1 The above sum could then be written, 1 0 0 0 0 1 0 0 +1 1 0 1 0 +2 0 0 0 1 0 0 1 We recover the matrix, and so we can write just as well, −~ω −b†b ~ω (2.15) ρ = 1 − exp i exp i i i e,i kT kT ! By the rules of probability, the density operator for the entire bath becomes, −~ω −b†b ~ω (2.16) ρ = 1 − exp i exp i i i e kT kT i ! Y By our assumptions, this is the state of the bath for all time, and so we will not write this as a function of time. The form ρs(t) will take will be much more complicated, and it is the purpose of this report to derive the differential equation defining its time evolution. 3 3. The Integration Beginning with Equation 2.10, we take the first integration to get, 1 t (3.1) ρ (t)= ρ (0) + [H (t′),ρ (t′)] dt′ se se i~ i se Z0 Taking this result, we plug it into itself and get, (3.2) ′ 1 t 1 t t ρ (t)= ρ (0)+ [H (t′),ρ (0)] dt′− [H (t′), [H (t′′),ρ (t′′)] dt′′dt′ se se i~ i se ~2 i i se Z0 Z0 Z0 Quoting Narducci, “now we do something unexpected” and differentiate this equa- tion. Since, dρ 1 se | = [H (0),ρ (0)] dt t=0 i~ i se and 0 anything = 0 Z0 we get the result, ρ (t) 1 1 t (3.3) se = [H (t),ρ (0)] − [H (t), [H (t′),ρ (t′)]] dt i~ i se ~2 i i se Z0 4. The Trace To extract ρs we now need to trace over the environmental variables, i.e. average out the environmental degrees of freedom. We can most quickly see that, (4.1) T re[Hi(t),ρse]=0 † as Hi is linear with b and bj , ρse = ρsrhoe, and it is self evident that ~ † T rej bj exp(− ωj bjbj /kT )=0 Thus we are left with, dρ (t) 1 t (4.2) s = − T r [H (t), [H (t′),ρ (t′)]] dt′ dt ~2 e i i se Z0 To proceed from here will require some care. 4.1. The partition. We first note that ′ ′ ′ ′ ′ ′ [Hi(t ),ρse(t )] = Hi(t )ρse(t ) − ρse(t )Hi(t ) ′ ′ which we use to write [Hi(t), [Hi(t ),ρse(t )]] = A C ′ ′ ′ ′ ′ ′ ′ ′ Hi(t)Hi(t )ρse(t ) − Hi(t)ρse(t )Hi(t ) − Hi(t )ρse(t )Hi(t)+ ρse(t )Hi(t )Hi(t) z }| { B z }| { D We will now proceed to| treat the{z integration} case by case. | {z } 4 4.2. Case A. ′ ′ Hi(t)Hi(t )ρse(t )= −~ω −b†b ~ω ~2 G(t)a† + G†(t)a G(t′)a† + G†(t′)a ρ (t′) 1 − exp i exp i i i s kT kT i ! Y The first term we can extract is, −~ω −b†b ~ω ~2 G(t)a†G†(t′)a ρ (t′) 1 − exp i exp i i i s kT kT i ! Y We recall that G(t)= gj bj exp(i(ω − ωj )t) j X ~ ~2 † ′ − ωi so we can rewrite the term as (with ζ = a aρs(t ) and χi = 1 − exp kT ), † ~ † ′ −bi bi ωi ζ gj bj exp(i(ω − ωj )t) gjb exp(−i(ω − ωj)t ) χi exp j kT j j i ! X X Y † It is easily seen that when we take the trace, all terms will involve bjbk and will vanish unless j = k, so we can write the term as, † ~ 2 † ′ −bi bi ωi ζ g bjb exp(i(ω − ωj )(t − t )) χi exp j j kT j i ! X Y But again, we can use the same argument to write the term as, † −b bj ~ωj ζ χ g2b b† exp(i(ω − ω )(t − t′)) exp j j j j j j kT j ! X We can now easily take the trace, noting that T r[Bρ] =<B>, and with n = b†b with [b,b†]=1.
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