3.4 Solutions to Problems
71
3.4.3 Derivation of χ (2)
[Problem 3.3] Derive ρ (2) (t) in Eq. (3.28).
Equation (3.27) is expanded by substituting ρ (1) with Eq. (3.22).
• Left side of Eq. (3.27):
dρ
(2)
mn
dt
+ iω mn ρ
(2)
mn + mn ρ
(2)
mn = e
−(iω mn + mn )t d
dt
e
(iω mn + mn )t ρ
(2)
mn (t)
• Right side of Eq. (3.27):
−
i
¯
h
H
, ρ
(1)
mn
=
i
¯
h
l
x−z
p
m|μ p |l
ρ
(1)
ln (t) − ρ
(1)
ml (t)
l|μ p |n
j
E p (ω j )e
−iω j t
=
i
¯
h 2
l
x−z
p,q
j,k
m|μ p |l
(ρ
(0)
l − ρ
(0)
n )
l|μ q |n
1
ω k − ω ln + ii ln
−(ρ
(0)
m − ρ
(0)
l )
m|μ q |l
1
ω k − ω ml + ii ml
l|μ p |n
E p (ω j )E q (ω k )e
−i(ω j +ω k )t .
Therefore,
d
dt
e
(iω mn + mn )t ρ
(2)
mn (t)
=
i
¯
h 2 e
(iω mn + mn )t
l
x−z
p,q
j,k
m|μ p |l
(ρ
(0)
l − ρ
(0)
n )
l|μ q |n
1
ω k − ω ln + ii ln
−(ρ
(0)
m − ρ
(0)
l )
m|μ q |l
1
ω k − ω ml + ii ml
l|μ p |n
E p (ω j )E q (ω k )e
−i(ω j +ω k )t .
This equation can be integrated by t from −∞ to t, using the following relation,
t
−∞
e
(iω mn + mn )τ e
−i(ω j +ω k )τ dτ =
e (iω mn + mn )t e −i(ω j +ω k )t
−i(ω j + ω k ) + iω mn + mn
(( mn > 0).