(
∂F
∂ρ−κF
ρ)
(
∂G
∂ρ+κG
ρ)
=
(√
k 2
k 1 −Zα
ρ)
G
(√
k 1
k 2 +Zα
ρ)
F
(
∂
∂ρ−κ
ρ)
F−
(√
k 2
k 1 −Zα
ρ)
G
(
∂
∂ρ+κ
ρ)
G−
(√
k 1
k 2 +Zα
ρ)
F
= 0
With the guidance of the non-relativistic solutions, we willpostulate a solution of the form
F=e−ρρs∑∞
m=0amρm=e−ρ∑∞
m=0amρs+mG=e−ρρs∑∞
m=0bmρm=e−ρ∑∞
m=0bmρs+m.The exponential will make everything go to zero for largeρif the power series terminates. We need
to verify that this is a solution nearρ= 0 if we pick the righta 0 ,b 0 , ands. We now substitute these
postulated solutions into the equations toobtain recursion relations.
(
∂
∂ρ−
κ
ρ)
F−
(√
k 2
k 1−
Zα
ρ)
G= 0
(
∂
∂ρ+
κ
ρ)
G−
(√
k 1
k 2+
Zα
ρ)
F= 0
∑∞
m=0(
−amρs+m+am(s+m)ρs+m−^1 −amκρs+m−^1 −bm√
k 2
k 1ρs+m+bmZαρs+m−^1)
= 0
∑∞
m=0(
−bmρs+m+bm(s+m)ρs+m−^1 +bmκρs+m−^1 −am√
k 1
k 2ρs+m−amZαρs+m−^1)
= 0
(
−am+am+1(s+m+ 1)−am+1κ−bm√
k 2
k 1+bm+1Zα)
= 0
(
−bm+bm+1(s+m+ 1) +bm+1κ−am√
k 1
k 2−am+1Zα)
= 0
−am+ (s+m+ 1−κ)am+1−√
k 2
k 1
bm+Zαbm+1= 0−bm+ (s+m+ 1 +κ)bm+1−√
k 1
k 2am−Zαam+1= 0−am+ (s+m+ 1−κ)am+1−√
k 2
k 1bm+Zαbm+1= 0−√
k 1
k 2am−Zαam+1−bm+ (s+m+ 1 +κ)bm+1= 0For the lowest order termρs, we need to have a solution without lower powers. This means that we