3.3 Solutions
199
When m = 0, (2) can be written as
∂
2
∂r 2 +
2
r
∂
∂r
−
m
2 c
2
2
ϕ(r ) = 0
or
1
r 2
∂
∂r
r
2 ∂ϕ
∂r
=
m
2 c
2
ϕ
2
(4)
For values of r > 0 from a point source at the origin, r = 0. Integration gives
ϕ(r ) =
ge
r
R
4πr
(5)
where R = /mc
(6)
The quantity g plays the same role as charge in electrostatistics and measures the “strong nuclear charge”.
3.3.4 Simple Harmonic Oscillator
3.51 By substituting ψ(R) = AH (R) exp (−R
2
/2) in the dimensionless form of
the equation and simplifying we easily get the Hermite’s equation
The problem is solved by the series method
H = ΣH n (R) = Σ n=0,2,4 a n R
n
dH
dR
= a n n R
n−1
d
2 H
dR 2 = Σn(n − 1)a n R
n−2
Σn(n − 1)a n R
n−2
− 2Σa n n R
n
+ (ε − 1)Σa n R
n
= 0
Equating equal power of R
n
a n+2 =
[2n − (ε − 1)] a n
(n + 1)(n + 2)
If the series is to terminate for some value of n then
2n − (ε − 1) = 0 becuase a n = 0. This gives ε = 2n + 1
Thus ε is a simple function of n
E = εE 0 = (2n + 1)
1
/ 2 ω, n = 0, 2, 4, . . .
=
1
/ 2 ω, 3ω/2, 5ω/2, . . .
Thus energy levels are equally spaced.