1000 Solved Problems in Modern Physics

168 3 Quantum Mechanics – II

(

1

sinθ

)

∂

∂θ

(

sinθ

df

dθ

)

+λf(θ)=0(2)

(

d

dθ

)

=

(

d

dμ

)

·

(

dμ

dθ

)

=−sinθ

d

dμ

Writingf(θ)=P(μ), Eq. (2) becomes

d

dμ

[

(

1 −μ^2

)dp

dμ

]

+λP= 0

or

(

1 −μ^2

)d^2 p

dμ^2

− 2 μ

dp

dμ

+λp=0(3)

One can solve Eq. (3) by series method

LetP=Σ∞k= 1 akμk (4)

dp

dμ

=

∑

k

akkμk−^1 (5)

d^2 P

dμ^2

=

∑

akk(k−1)μk−^2 (6)

Using (4), (5) and (6) in (3)

∑

k(k−1)akμk−^2 −

∑

k(k−1)akμk− 2

∑

kakμk+λΣakμk= 0

Equating equal powers ofk

(k+2)(k+1)ak+ 2 −[k(k−1)+ 2 k−λ]ak= 0

Orak+ 2 /ak=[k(k+ 1 )−λ]/(k+ 1 )(k+ 2 )

(b) If the infinite series is not terminated, it will diverge atμ=±1, i.e. at

θ =0orθ=π. Because this should not happen the series needs to be

terminated which is possible only ifλ=k(k+1)

i.e.l(l+1);l= 0 , 1 , 2 ...Herelis known as the orbital angular momen-

tum quantum number. The resulting seriesP(μ) is then called Legendre

polynomial.

3.3.3 PotentialWellsandBarriers .........................

3.18 (a) The term−

(^2) d 2
2 mdx^2 is the kinetic energy operator,U(x) is the potential
energy operator,ψ(x) is the eigen function andEis the eigen value.
(b) PutU(x)=0 in the region 0<x <ain the Schrodinger equation to
obtain
(
−

^2

2 m

)

d^2 ψ(x)

dx^2

=Eψ(x)(1)