586 Langevin and generalized Langevin equations
ω ̃^2 =ω^2 − 2∑
αg^2 α
μmαω^2 α, (15.3.19)
so that
W(q) =1
2
μω ̃^2 q^2. (15.3.20)The quantity ̃ωis known as therenormalized frequencyWe will examine the case
in which the frequency of the oscillator is high compared to the bath frequencies, a
condition that exists when the coupling between the system and thebath is weak. In
the limit of high ̃ω, the term− 2
∑
αg
2
α/μmαω
2
αwill be a small perturbation toω(^2).
For a general friction kernelζ(t), the GLE reads
q ̈=−ω ̃^2 q−
∫t
0
dτq ̇(τ)γ(t−τ) +f(t), (15.3.21)
whereγ(t) =ζ(t)/μandf(t) =R(t)/μ. Eqn. (15.3.21) must be solved subject to initial
conditionsq(0) and ̇q(0). Taking the Laplace transform of both sides and solving for
q ̃(s) yields
q ̃(s) =
(s+ ̃γ(s))
∆(s)
q(0) +
q ̇(0)
∆(s)
+
f ̃(s)
∆(s), (15.3.22)
where
∆(s) =s^2 + ̃ω^2 +s ̃γ(s). (15.3.23)
In order to perform the Laplace inversion, the poles of each of theterms on the right
side of eqn. (15.3.22) are needed. These are given by the zeroes of∆(s). That is, we
seek solutions of
s^2 + ̃ω^2 +s ̃γ(s) = 0. (15.3.24)
Even if we do not know the explicit form of ̃γ(s), when ̃ωis large compared to the
bath frequencies, it is possible to solve eqn. (15.3.24) perturbatively. We do this by
positing a solution to eqn. (15.3.24) forsof the form
s=s 0 +s 1 +s 2 +··· (15.3.25)as an ansatz (Tuckerman and Berne, 1993). Substituting eqn. (15.3.25) into eqn.
(15.3.24) gives
(s 0 +s 1 +s 2 +···)^2 + ̃ω^2 + (s 0 +s 1 +s 2 +···) ̃γ(s 0 +s 1 +s 2 ···) = 0. (15.3.26)Assuming ̃ω^2 >> sγ ̃(s) at the root, we can solve this equation to lowest order by
neglecting thes ̃γ(s) term, which gives
s^20 + ̃ω^2 = 0, s 0 =±iω. ̃ (15.3.27)Next, working to first order in the perturbation, we have
s^20 + 2s 0 s 1 + ̃ω^2 +s 0 γ ̃(s 0 ) = 0, (15.3.28)