130_notes.dvi

We must use the Hankel function of the first type for larger.

ρ=kr→iκr

κ=

√

− 2 μE

̄h^2

Rnℓ=Bh(1)ℓ (iκr)

To solve the problem, we have to match the solutions at the boundary. First match the wavefunction.

A[jℓ(ρ)]ρ=ka=B[hℓ(ρ)]ρ=iκa

Then match the first derivative.

Ak

[

djℓ(ρ)

dρ

]

ρ=ka

=B(iκ)

[

dhℓ(ρ)

dρ

]

ρ=iκa

We can divide the two equations to eliminate the constants to get a condition on the energies.

k

[djℓ(ρ)

dρ

jℓ(ρ)

]

ρ=ka

= (iκ)

[dhℓ(ρ)

dρ

hℓ(ρ)

]

ρ=iκa

This is often called matching thelogarithmic derivative.

Often, theℓ= 0 term will be sufficient to describe scattering. Forℓ= 0, the boundary condition is

k

[cosρ

ρ −

sinρ

ρ^2

sinρ

ρ

]

ρ=ka

= (iκ)

[ieiρ

iρ −

eiρ

iρ^2

eiρ

iρ

]

ρ=iκa

.

Dividing and substituting forρ, we get

k

(

cot(ka)−

1

ka

)

=iκ

(

i−

1

iκa

)

.

kacot(ka)−1 =−κa− 1

kcot(ka) =−κ

This is thesame transcendental equation that we had for the odd solution in one dimen-
sion.

−cot





√

2 μ(E+V 0 )

̄h^2

a



=

√

−E

V 0 +E

The number of solutions depends on the depth and radius of the well.There can even be no solution.