1000 Solved Problems in Modern Physics

3.3 Solutions 199

Whenm

=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 ofr>0 from a point source at the origin,r=0. Integra-

tion gives

φ(r)=

ge

rR

4 πr

(5)

whereR=/mc (6)

The quantitygplays the same role as charge in electrostatistics and mea-

sures the “strong nuclear charge”.

3.3.4 Simple Harmonic Oscillator .........................

3.51 By substitutingψ(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=ΣHn(R)=Σn= 0 , 2 , 4 anRn

dH

dR

=annRn−^1

d^2 H

dR^2

=Σn(n−1)anRn−^2

Σn(n−1)anRn−^2 − 2 ΣannRn+(ε−1)ΣanRn= 0

Equating equal power ofRn

an+ 2 =

[ 2 n−(ε−1)]an

(n+1)(n+2)

If the series is to terminate for some value ofnthen

2 n−(ε−1)=0 becuasean = 0 .This givesε= 2 n+ 1

Thusεis a simple function ofn

E=εE 0 =(2n+1)^1 / 2 ω,n= 0 , 2 , 4 ,...

=^1 / 2 ω, 3 ω/ 2 , 5 ω/ 2 ,...

Thus energy levels are equally spaced.