1000 Solved Problems in Modern Physics

166 3 Quantum Mechanics – II

Following the same procedure

g(φ)

sinθ

d

dθ

sinθ

df(θ)

dθ

+

f(θ)

sin^2 θ

d^2 g(φ)

dφ^2

+λf(θ)g(φ)= 0

sinθ

f(θ)

d

dθ

(

sinθ

df(θ)

dθ

)

+λsin^2 θ=

− 1

g(φ)

d^2 g(φ)

dφ^2

=m^2 (7)

wherem^2 is a positive constant

d^2 g

dφ^2

=−m^2 φ (8)

gives the normalized function

g=(1/

√

2 π)eimφ (9)

mis an integer sinceg(φ+ 2 π)=g(φ)

Dividing (6) by sin^2 θand multiplying byf, and rearranging

1

sinθ

d

dθ

(

sinθ

df

dθ

)

+

(

λ−

m^2

sin^2 θ

)

f= 0 (10)

(c) The physically accepted solution of (10) is Legendre polynomials when

λ=l(l+1) (11)

andlis an integer.

With the change of variableψr(r)=χ(r)/r

The first term in (5) becomes

d

dr

(

r^2

ψr

dr

)

=

d

dr

[

r^2

(

−

χ

r^2

+

1

r

dχ

dr

)]

=

d

dr

[

r

dχ

dr

−χ

]

=r

d^2 χ

dr^2

+

dχ

dr

−

dχ

dr

=r

d^2 χ

dr^2

With the substitution ofλfrom (11), (5) becomes upon rearrangement

(

−

^2

2 m

)

d^2 χ

dr^2

+

[

V(r)+

l(l+1)^2

2 mr^2

]

χ=Eχ (12)

Thus, the radial motion is similar to one dimensional motion of a particle

in a potential

Ve=V(r)+

l(l+1)^2

2 mr^2

(13)

whereVeis the effective potential. The additional “potential energy” is

interpreted to arise physically from the angular momentum. A classical

particle that has angular momentumLabout the axis through the origin

perpendicular to the plane of its path has the angular velocityω=L/mr^2

where its radial distance from the origin isr. An inward forcemω^2 r=

mL^2 /ωr^3 is required to keep the particle in the path. This “centripetal