14.3 Echoes 423
from thebpulse in the second ̃φextfactor in Eq.(14.135) will be considered. We therefore
substitute theacontribution in Eq.(14.137) for the first ̃φextfactor in Eq.(14.135) to obtain
̃φ 2 (p,k) = ωpπφa
k^2 D(p,k)
e
m
∫
dv
∫∞
−∞
dk′
2 π
∫b+i∞
b−i∞
dp′
2 π
ikk′δ(k′±ka)
(p+ikv)^2 D(p′,k′)
×
i ̄k′
(p−p′)+i ̄k′v
̃φext((p−p′), ̄k′)
D(p−p′, ̄k′)
∂f ̄ 0
∂v
. (14.138)
The effect of theδ(k′±ka)factor when evaluating thek′integral is to forcek′→±ka
and also ̄k′→k∓kaso that
φ ̃ 2 (p,k) = ωp
k^2 D(p,k)
eφa
2 m
∑
±
∫
dv
∫b+i∞
b−i∞
dp′
2 π
(∓ikka)
(p+ikv)^2 D(p′,∓ka)
×
i (k∓ka)
p ̄′+i(k ∓ka)v
φ ̃ext( ̄p′,(k∓ka))
D( ̄p′,k∓ka)
∂f ̄ 0
∂v
. (14.139)
Now let us substitute for the secondφ ̃extfactor using the contribution from thebpulse to
obtain
φ ̃ 2 (p,k) = π
2 k^2 D(p,k)
e
m
φaφb
∑
±
∑
±
∫
dv
∫b+i∞
b−i∞
dp′
2 π
(∓ikka)
(p+ikv)^2 D(p′,∓ka)
×
i (k∓ka)
p ̄′+i(k ∓ka)v
δ(k∓ka±kb)e−p ̄
′τ
D( ̄p′,k∓ka)
∂f ̄ 0
∂v
. (14.140)
The upper choice of the±and∓signs is selected and the inverse Fourier transform per-
formed to obtain
̃φupper 2 (p,x) = φaφb
4 k^2 D(p,k)
e
m
∫∞
−∞
dk
∫
dv
∫b+i∞
b−i∞
dp′
2 π
(−ikka)
(p+ikv)^2 D(p′,−ka)
×
i (k−ka)
p ̄′+i(k−ka)v
δ(k−ka+kb)e− ̄p
′τ+ikx
D( ̄p′,k−ka)
∂f ̄ 0
∂v
=
φaφb
4(ka−kb)^2 D(p,ka−kb)
e
m
∫
dv
∫b+i∞
b−i∞
dp′
2 π
×
i(ka−kb)
(p+i(ka−kb)v)^2
ka
D(p′,−ka)
ikb
p ̄′−ikbv
e− ̄p
′τ+i(ka−kb)x
D( ̄p′,−kb)
∂f ̄ 0
∂v
.
(14.141)
The lower choice of the±and∓signs means thatka→−kaandkb→−kband so at
the end of the calculationφ ̃
lower
2 (p,x)can also be determined by simply lettingka→−ka