QuantumPhysics.dvi

Now, we know the solutionjℓ(kr) for U = 0, and consider the full solution in a perturbative
expansion inU. We set

Rℓ(r) =jℓ(kr) +χℓ(r) (12.76)

To solve this problem, we define the Green function for the 1-dimensional differential operator as
follows,
(
d^2
dr^2

+

2

r

d

dr

−

ℓ(ℓ+ 1)

r^2

+k^2

)

Gℓ(r,r′) =

1

r^2

δ(r−r′) (12.77)

The reason for the factor of 1/r^2 on the rhs is that the integration measure isd^3 r=r^2 drsinθdθdφ,
and theδ-function in spherical coordinates thus inherits a compensating 1/r^2 factor. In terms of
this Green function, the integral equation forRℓbecomes,

Rℓ(r) =jℓ(kr) +

∫∞

0

dr′(r′)^2 Gℓ(r,r′)U(r′)Rℓ(r′) (12.78)

This integral equation may again be solved recursively.

12.10.3Calculating the radial Green function

There is a standard, and very powerful, method for calculating the Green function for linear differ-
ential equations of second order in one variable. The methodproceeds a s follows. Whenr 6 =r′, the
equation is just the homogeneous one, for which we know the two linearly independent solutions
jℓ(kr) andhℓ(kr). Atr=r′, the presence of aδ-function requires the derivative inrofGℓ(r,r′) to
be discontinuous, and thus the functionGℓ(r,r′) itself to be continuous. This may be achieved by
using different linear combinations of the functionsjℓandhℓforr < r′and forr > r′.

We shall choose these linear combinations such thatGℓ(r,r′) is automatically regular atr= 0,
and has the correcteikrbehavior asr→ ∞. The functionjℓ(kr) is regular atr= 0, while the
functionhℓ(kr) behaves likeeikrasr→∞. Therefore, we use the following form forGℓ(r,r′),

Gℓ(r,r′) =

{

Cℓjℓ(kr)hℓ(kr′) r < r′

Cℓhℓ(kr)jℓ(kr′) r′< r

(12.79)

This expression guarantees continuity acrossr=r′from the outset. We shall now recast this form
in a more helpful form using the Heaviside step functionθ(x). It is defined by

θ(x) =

{

1 x > 0

0 x < 0

(12.80)

Its derivative is theδ-function,θ′(x) =δ(x). Using the Heaviside function, we now have

Gℓ(r,r′) =Cℓθ(r′−r)jℓ(kr)hℓ(kr′) +Cℓθ(r−r′)hℓ(kr)jℓ(kr′) (12.81)

In differentiating once, continuity ofGℓ(r,r′) acrossr=r′guarantees that the resultingδ-functions
will cancel, and we obtain,

d

dr

Gℓ(r,r′) =kCℓθ(r′−r)jℓ′(kr)hℓ(kr′) +kCℓθ(r−r′)h′ℓ(kr)jℓ(kr′) (12.82)