130_notes.dvi

We nowtake the transform of both sidesof the equation.

∇^2 φ−μ^2 φ = Gδ^3 (~x)

1

(2π)

(^32)

∫

d^3 x e−i

~k·~x

(∇^2 φ−μ^2 φ) =

1

(2π)

(^32)

∫

d^3 x e−i

~k·~x

Gδ^3 (~x)

1

(2π)

(^32)

∫

d^3 x e−i

~k·~x

(∇^2 φ−μ^2 φ) =

G

(2π)

(^32)
1
(2π)
(^32)

∫

d^3 x e−i

~k·~x

(−k^2 −μ^2 )φ =

G

(2π)

(^32)
(−k^2 −μ^2 )φ ̃ =

G

(2π)

(^32)
φ ̃ = −G
(2π)
(^32)

1

(k^2 +μ^2 )

To deal with the∇^2 , we have integrated by parts twice assuming that the field falls off fast enough
at infinity.

We now have theFourier transform of the field of a point source. If we cantransform back
to position space, we will have the field. This is a fairly standard type of problem in quantum
mechanics.

φ(~x) =

1

(2π)

(^32)

∫

d^3 k ei

~k·~x −G

(2π)

(^32)

1

(k^2 +μ^2 )

=

−G

(2π)^3

∫

d^3 k

ei~k·~x

(k^2 +μ^2 )

=

− 2 πG

(2π)^3

∫

k^2 dk

∫^1

− 1

eikrcosθk

(k^2 +μ^2 )

dcosθk

=

−G

(2π)^2

∫

k^2

(k^2 +μ^2 )

∫^1

− 1

eikrcosθkdcosθkdk

=

−G

(2π)^2

∫

k^2

(k^2 +μ^2 )

[

1

ikr

eikrcosθk

] 1

− 1

dk

=

−G

(2π)^2 ir

∫

k

(k^2 +μ^2 )

(

eikr−e−ikr

)

dk

=

−G

(2π)^2 ir

∫∞

0

k

(k+iμ)(k−iμ)

(

eikr−e−ikr

)

dk

This is now of a form for which we can use Cauchy’s theorem forcontour integrals. The theorem
says that anintegral around a closed contour in the complex planeis equal to 2πitimes
the sum of the residues at the poles enclosed in the contour. Contour Integration is a powerful
technique often used in quantum mechanics.

If the integrand is a function of the complex variablek, a pole is of the formrkwhereris called the
residue at the pole. Theintegrand above has two poles, one atk=iμand the other atk=−iμ.
Theintegral we are interested in is just along the real axisso we want the integral along