QuantumPhysics.dvi

The factor of 4πgives the correct normalization for theδ-function. Fork^26 = 0, rotation invariance
of the Fourier integral shows thatG±depends only on the scalarsk^2 andr, so that, away from
x=y,G±must satisfy the radial equation

1

r^2

∂

∂r

(

r^2

∂G±

∂r

)

+k^2 G±= 0 (12.18)

wherek=|k|. Using dimensional analysis, we must haveG±(x,y;k^2 ) =G±(x,y; 0)f(kr), where
f(kr) is a dimensionless function of the dimensionless combinationkr. Substitution into the above
radial differential equations shows that

f′′(kr) +f(kr) = 0 (12.19)

whose solutions aref(kr) =e±ikr. Putting all together, we obtain,

G±(x,y;k^2 ) =−

e±ik|x−y|

4 π|x−y|

(12.20)

Note that, since the exponentials equal 1 whenx=y, this result is automatically properly nor-
malized with respect to theδ-function source.

12.4 The Lippmann-Schwinger equation in position space

In many cases of interest, the perturbing HamiltonianH 1 islocal in position spaceand actually
given by a potentialV(x),

〈x|H 1 |y〉=δ(3)(x−y)V(x) (12.21)

In terms of the incoming wave functionφk(x) and the full wave functionψk(x), defined by

〈x|φk〉=φk(x) 〈x|ψ±k〉=ψ±k(x) (12.22)

the Lippmann-Schwinger equation takes the form,

ψ±k(x) =φk(x) +

∫

d^3 yG±(x,y;k^2 )U(y)ψ±k(y) (12.23)

where we have used the abbreviation

U(y) =

2 mλ

̄h^2

V(y) (12.24)

so thatUis automatically of orderλ, and thus small. If we symbolically denote this notation by

ψ=φ+GUψ (12.25)

then the solution may be obtained in a power series inλby iterating the equation, i.e. substituting
the left hand side into the right hand side repeatedly. At thefirst iteration, we obtain,

ψ=φ+GUφ+GUGUψ (12.26)