Stochastic Methods

Abstract In all physical processes there is an associated loss mechanism. In this chapter we shall consider how losses may be included in the quantum mechanical equations of motion. There are several ways in which a quantum theory of damping may be developed. We shall adopt the following approach: We consider the system of interest coupled to a heat bath or reservoir. We first derive an operator master equation for the density operator of the system in the Schro¨dinger or interaction picture. Equations of motion for the expectation values of system operators may directly be derived from the operator master equation. Using the quasi-probability representations for the density operator discussed in Chap. 4, the operator master equation may be converted to a c-number Fokker–Planck equation. For linear prob- lems a time-dependent solution to the Fokker–Planck equation may be found. In cer- tain nonlinear problems with an appropriate choice of representation the steady-state solution for the quasi-probability distribution may be found from which moments may be calculated.
Using methods familiar in stochastic processes the Fokker–Planck equation
may be converted into an equivalent set of stochastic differential equations. These stochastic differential equations of which the Langevin equations are one example are convenient when linearization is necessary. We begin then with a derivation of the master equation. We follow the method of Haake [1].



6.1 Master Equation


We consider a system described by the Hamiltonian HS coupled to a reservoir de- scribed by the Hamiltonian HR. The reservoir may be considered to be a large num- ber of harmonic oscillators as, for example, the modes of the free electromagnetic field or phonon modes in a solid. In some cases the reservoir may be more appro- priately modelled as a set of atomic energy levels. The derivation of the master equation is not dependent on the specific reservoir model. There is a weak inter- action between the system and the reservoir given by the Hamiltonian V . Thus the total Hamiltonian is


93


H = HS + HR + V (6.1)
Let w(t) be the total density operator of the system plus reservoir in the interaction picture. The equation of motion in the interaction picture is


dw(t) dt


i
= h¯ [V (t), w(t)] (6.2)

The reduced density operator for the system is defined by
? (t)= TrR{w(t)} (6.3)
where TrR indicates a trace over reservoir variables. We assume that initially the system and reservoir are uncorrelated so that
w(0)= ? (0) ? ?R (6.4)

where ?R is the density operator for the reservoir.
Integrating (6.2) we obtain


t
i ¸
w(t)= w(0) ? h¯
0



dt1[V (t1), w(t1 )] . (6.5)

Iterating this solution we find



? . i .n ¸t t1

w(t)= w(0)+ ?
n=1

? h¯

dt1
0 0

dt2 ···

tn?1
¸
× dtn[V (t1), [V (t2),... [V (tn), w(0)]]] . (6.6)
0

Performing the trace over reservoir variables

? . i .n ¸t t1


tn?1
¸

? (t)= ? (0)+ ?
n=1

? h¯

dt1
0 0

dt2 ···
0

dtn






where

× TrR{[V (t1), [V (t2),... [V (tn), ?R ? ? (0)]]]}
? (1 + U1(t)+ U2(t)+ ··· )? (0) (6.7)
? U (t)? (0)

. i .n

t t1
¸ ¸

Un(t)=

? h¯

tn?1
¸

TrR
0

dt1
0

dt2 ···

× dtn[V (t1), [V (t2),... [V (tn), ?R ? (·)]] .. .] . (6.8)
0



Thus



= [U?1(t)+ U?2(t)+ ··· ]U (t)?1? (t)
? l(t)? (t) (6.9)


where the generator of time development is
l(t)= [U?1(t)+ U?2(t)+ ··· ]U (t)?1 . (6.10)

We now assume that V (t) is such that

TrR(V (t)?R)= 0 . (6.11)

This ensures that U1(t)= 0. If the perturbation is weak we may drop terms from l(t)
of order higher than two. Thus

l(t)= U?2(t)

t
1 ¸
= ? h¯ 2
0


dt1 TrR[V (t), [V (t1), ?R ? (·)]] . (6.12)


Thus to second order in the perturbation



d? =
dt

t
1 ¸
? h¯ 2
0



dt1



TrR


[V (t), [V (t1), ?R ?



? (t)]] . (6.13)


The next-order correction is at least quartic in the coupling and thus we expect (6.13) to be a good approximation.
Let us now consider the case of a damped simple harmonic oscillator. In this case

V (t)= h¯ (a†?(t)ei?0t + a?†(t)e?i?0t ) (6.14)


where



and


?(t)= ?g j b je?i? jt , (6.15)
j


[b j , b†]= ? jk . (6.16)


Substituting (6.14) into (6.13) we find that the following integrals are required


t
¸
I1 =
0
t
¸
I2 =
0


dt1(?(t)?(t1))ei?0 (t+t1 ) , (6.17)


dt1(?†(t)?†(t1))e?i?0(t+t1 ) , (6.18)










which we now evaluate.


t
¸
I3 =
0
t
¸
I4 =
0


dt1(?(t)?†(t1))ei?0 (t?t1 ) , (6.19)


dt1(?†(t)?(t1))e?i?0 (t?t1 ) (6.20)

Using the definition of ?(t) we have

t

¸
I1 =
0

dt1 ?gig j(bib j )Re?i(?it+? jt1 )ei?0(t+t1 ) . (6.21)
i, j


Converting the sum over modes to a frequency-space integral


t
¸
I1 =
0


?
¸
dt1
0


?
d?1 ¸
2? ? (?1)
0


d?2
2? ? (?2)g(?1)g(?2)(b(?1)b(?2))R

× e?i(?1t+?2t1 )+i?0(t+t1 ) (6.22)