130_notes.dvi

Next we look at the equation forlarger.

d^2 R

dρ^2

−

1

4

R= 0

This can be solved byR=e

−ρ

(^2) , so we explicitly include this.
R(ρ) =G(ρ)e
−ρ
2
We should also pick of thesmallrbehavior.
d^2 R
dρ^2

+

2

ρ

dR

dρ

−

ℓ(ℓ+ 1)

ρ^2

R= 0

AssumingR=ρs, we get

s(s−1)

R

ρ^2

+ 2s

R

ρ^2

−ℓ(ℓ+ 1)

R

ρ^2

= 0.

s^2 −s+ 2s=ℓ(ℓ+ 1)

s(s+ 1) =ℓ(ℓ+ 1)

So eithers=ℓors=−ℓ−1. The second is not well normalizable. WewriteGas a sum.

G(ρ) =ρℓ

∑∞

k=0

akρk=

∑∞

k=0

akρk+ℓ

The differential equation forG(ρ) is

d^2 G

dρ^2

−

(

1 −

2

ρ

)

dG

dρ

+

(

λ− 1

ρ

−

ℓ(ℓ+ 1)

ρ^2

)

G(ρ) = 0.

We plug the sum into the differential equation.

∑∞

k=0

ak

(

(k+ℓ)(k+ℓ−1)ρk+ℓ−^2 −(k+ℓ)ρk+ℓ−^1 + 2(k+ℓ)ρk+ℓ−^2

+(λ−1)ρk+ℓ−^1 −ℓ(ℓ+ 1)ρk+ℓ−^2

)

= 0

∑∞

k=0

ak((k+ℓ)(k+ℓ−1) + 2(k+ℓ)−ℓ(ℓ+ 1))ρk+ℓ−^2

=

∑∞

k=0

ak((k+ℓ)−(λ−1))ρk+ℓ−^1

Now weshift the sumso that each term containsρk+ℓ−^1.

∑∞

k=− 1

ak+1((k+ℓ+ 1)(k+ℓ) + 2(k+ℓ+ 1)−ℓ(ℓ+ 1))ρk+ℓ−^1 =

∑∞

k=0

ak((k+ℓ)−(λ−1))ρk+ℓ−^1