as one expects since the delta-functionδ(ω) is the Fourier transform of
1
√
2 π
(θ(t) +θ(−t)) =
1
√
2 π
Returning to the propagator, as in section 11.5 one can Fourier transform
with respect to time, and thus get a propagator that depends on the frequency
ω. The Fourier transform of equation 12.6 with respect to time is
Û(ω,k) =√^1
2 π
∫+∞
−∞
(
1
√
2 π
e−i
21 mk^2 t
)
eiωtdt=δ(ω−
1
2 m
k^2 )
Using equations 12.5 and 12.10 the retarded progagator in position space is
given by
U+(t,qt−q 0 ) = lim
→ 0 +
(
1
2 π
) 2 ∫+∞
−∞
∫+∞
−∞
i
ω+i
e−iωteik(qt−q^0 )e−i
21 mk^2 t
dωdk
= lim
→ 0 +
(
1
2 π
) 2 ∫+∞
−∞
∫+∞
−∞
i
ω+i
e−i(ω+
21 mk^2 )t
eik(qt−q^0 )dωdk
Shifting the integration variable by
ω→ω′=ω+
1
2 m
k^2
one finds
U+(t,qt−q 0 ) = lim
→ 0 +
(
1
2 π
) 2 ∫+∞
−∞
∫+∞
−∞
i
ω′− 21 mk^2 +i
e−iω
′t
eik(qt−q^0 )dω′dk
but this is the Fourier transform
U+(t,qt−q 0 ) =
1
2 π
∫+∞
−∞
∫+∞
−∞
Û+(ω,k)e−iωteik(qt−q^0 )dωdk (12.12)
where
Û+(ω,k) = lim
→ 0 +
i
2 π
1
ω− 21 mk^2 +i
(12.13)
Digression.By the same argument as the one above for the integral represen-
tation ofθ(t), but with pole now at
ω=
1
2 m
k^2 −i
theωintegral in equation 12.12 can be evaluated by the Cauchy integral formula,
recovering formula 12.9 forU(t,qt−q 0 ).