Quantum Mechanics for Mathematicians

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 ).