1000 Solved Problems in Modern Physics

(Romina) #1

168 3 Quantum Mechanics – II


(
1
sinθ

)


∂θ

(

sinθ

df

)

+λf(θ)=0(2)
(
d

)

=

(

d

)

·

(



)

=−sinθ

d

Writingf(θ)=P(μ), Eq. (2) becomes
d

[

(

1 −μ^2

)dp

]

+λP= 0

or

(

1 −μ^2

)d^2 p
dμ^2

− 2 μ

dp

+λp=0(3)

One can solve Eq. (3) by series method
LetP=Σ∞k= 1 akμk (4)
dp

=


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)
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