1000 Solved Problems in Modern Physics

(Romina) #1

166 3 Quantum Mechanics – II


Following the same procedure
g(φ)
sinθ

d

sinθ

df(θ)

+

f(θ)
sin^2 θ

d^2 g(φ)
dφ^2

+λf(θ)g(φ)= 0

sinθ
f(θ)

d

(

sinθ

df(θ)

)

+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

(

sinθ

df

)

+

(

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


dr

)]

=

d
dr

[

r


dr

−χ

]

=r

d^2 χ
dr^2

+


dr



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
Free download pdf