130_notes.dvi

where

ρ=

2 Z

na 0

r

and the coefficients come from the recursion relation

ak+1=

k+ℓ+ 1−n

(k+ 1)(k+ 2ℓ+ 2)

ak.

The series terminates fork=n−ℓ−1.

Lets start withR 10.

R 10 (r) =ρ^0

∑^0

k=0

akρke−ρ/^2

R 10 (r) =Ce−Zr/a^0

We determineCfrom the normalization condition.

∫∞

0

r^2 R∗nℓRnℓdr= 1

|C|^2

∫∞

0

r^2 e−^2 Zr/a^0 dr= 1

This can be integrated by parts twice.

2

(a

0

2 Z

) 2

|C|^2

∫∞

0

e−^2 Zr/a^0 dr= 1

2

(a

0

2 Z

) 3

|C|^2 = 1

C^2 =

1

2

(

2 Z

a 0

) 3

C=

1

√

2

(

2 Z

a 0

) (^32)
R 10 (r) = 2

(

Z

a 0

) (^32)
e−Zr/a^0
R 21 can be computed in a similar way. No recursion is needed.
Lets tryR 20.
R 20 (r) =ρ^0

∑^1

k=0

akρke−ρ/^2

R 20 (r) = (a 0 +a 1 ρ)e−ρ/^2