Examples 587where it has been assumed thats 0 ̃γ(s 0 ) is of the same order ass 1. Using the fact that
s^20 =−ω ̃^2 and solving fors 1 , we obtain
s 1 =−1
2
γ ̃(±iω ̃), (15.3.29)which requires the evaluation of ̃γ(s) ats=±iω ̃. Note that
γ ̃(±iω ̃) =∫∞
0dt γ(t)e∓iωt ̃, (15.3.30)which contains both real and imaginary parts. Defining
γ ̃(±iω ̃) =γ′( ̃ω)∓iγ′′( ̃ω), (15.3.31)there are, to first order, two roots of ∆(s), which are given by
s+=i(
ω ̃+1
2
γ′′( ̃ω))
−
1
2
γ′( ̃ω)≡iΩ−1
2
γ′( ̃ω)s−=−i(
ω ̃+1
2
γ′′( ̃ω))
−
1
2
γ′( ̃ω)≡−iΩ−1
2
γ′( ̃ω). (15.3.32)Substituting the roots in eqn. (15.3.32) into eqn. (15.3.22) yields thesolution
q(t) =q(0)e−γ′( ̃ω)t/ 2[
cos Ωt+γ′( ̃ω)
2Ωsin Ωt]
+
q ̇(0)
Ωe−γ′( ̃ω)t/ 2
sin Ωt+
1
Ω
∫t0dτ f(t−τ)e−γ′( ̃ω)τ/ 2
sin Ωτq ̇(t) = ̇q(0)[
cos Ωt−γ′( ̃ω)
2Ω
sin Ωt]
e−γ′( ̃ω)t/ 2
−Ωq(0)e−γ′( ̃ω)t/ 2
sin Ωt+
1
Ω
[
f(0)e−γ′( ̃ω)t/ 2
sin Ωt+∫t0dτ f′(t−τ)e−γ′( ̃ω)τ/ 2
sin Ωτ]
, (15.3.33)
where we have used the fact thatγ′′( ̃ω)≪ Ω, and we have neglected any terms
nonlinear inγ′( ̃ω). Since〈q(0)f(t)〉= 0 and〈q ̇(0)f(t)〉= 0, the velocity and position
autocorrelation functions become, respectively
Cvv(t) =〈
q ̇^2 (0)〉
e−γ′( ̃ω)t/ 2[
cos Ωt−γ′( ̃ω)
2Ωsin Ωt]
Cqq(t) =〈
q^2 (0)〉
e−γ′( ̃ω)t/ 2[
cos Ωt+
γ′( ̃ω)
2Ωsin Ωt]
. (15.3.34)
As eqn. (15.3.34) shows, the decay time of both correlation functions is [γ′( ̃ω)t/2]−^1 ,
which is denotedT 2 and is called thevibrational dephasing time. (We will explore