14.5 The particle in a circular box 399
or, multiplying throughout by− 2 mr
22A
2, setting
(14.33)
and rearranging,
(14.34)
This equation can now be reduced to two ordinary equations by writing the wave
function as the product
ψ(r, 1 θ) 1 = 1 R(r) 1 × 1 Θ(θ) (14.35)
Substitution in (14.34) and division byψ 1 = 1 RΘthen gives
(14.36)
Each set of terms in square brackets must be constant so that, with separation
constant C,
(14.37)
for the radial motion of the particle in the box, and
(14.38)
for the angular motion of the particle. We consider the angular equation first.
The angular equation
The functionΘ(θ)is defined in the interval 01 ≤ 1 θ 1 ≤ 12 πand must satisfy the periodic
boundary condition
Θ(2π) 1 = 1 Θ(0) (14.39)
for continuity round the circle. We therefore have the same boundary value problem
as that discussed in Section 12.7 for the particle in a ring. The normalized solutions
are (equation (12.69))
n 1 = 1 0, ±1, ±2, = (14.40)
and substitution in (14.38) gives the separation constantC 1 = 1 n
2.
Θ
nin()θ e
θ=,
1
2 π
d
d
C
22Θ
Θ
θ
=−
r
dR
dr
r
dR
dr
rR CR
22222++ =α
r
R
dR
dr
r
R
dR
dr
r
d
d
22222221
+++
α
θ
Θ
Θ
= 0
r
r
r
r
r
22222220
∂
∂
∂
∂
++
∂
∂
=
ψψ
αψ
ψ
θ
α
222
=
mE