1000 Solved Problems in Modern Physics

224 3 Quantum Mechanics – II

whereF(r)=rf(r)

Lzψ 1 =−i

∂

∂φ

(cosφ+isinφ)sinθF(r)

=−i(−sinφ+icosφ)sinθF(r)

=(cosφ+isinφ)sinθF(r)

=ψ 1

Thusψ 1 is the eigen state and the eigen value is

ψ 2 =zf(r)=rcosθf(r)

Lzψ 2 =−i

∂

∂φ

(rcosθf(r))= 0

The eigen value is zero.

ψ 3 =(x−iy)f(r)=rsinθ(cosφ−isinφ)f(r)

=(cosφ−isinφ)sinθF(r)

Lzψ 3 =−i

∂

∂φ

(cosφ−isinφ)sinθF(r)

=−i(−sinφ−icosφ)sinθF(r)

=i(sinφ+icosφ)sinθF(r)

=−(cosφ−isinφ)sinθF(r)=−ψ 3

Thusψ 3 is an eigen state and the eigen value is−.

3.89 (a)Lz=−i∂
∂φ

Lzψ=−i

∂

∂φ

Af(r)sinθcosθeiφ

=(i)(−iAf(r)sinθ cosθeiφ)

=Af(r)sinθ cosθeiφ

=ψ

Therefore, thez-component of the angular momentum is.

(b)L^2 =−^2

{

∂^2

∂θ^2 +cotθ

∂

∂θ+

(

1

sin^2 θ

)

∂^2

∂φ^2

}

Expressions forLzandL^2 are derived in Problems 3.80 and 3.83.

L^2 ψ=−^2

{

∂^2

∂θ^2

+cotθ

∂

∂θ

+

(

1

sin^2 θ

)

·

∂^2

∂φ^2

}

Af(r)sinθcosθeiφ

=−^2 Af(r)eiφ

{

−4sinθcosθ+cotθ(cos^2 θ−sin^2 θ)−

sinθcosθ

sin^2 θ

}

= 6 ^2 Af(r)sinθcosθeiφ

= 6 ^2 ψ

ThusL^2 = 6 ^2

ButL^2 =l(l+1). Thereforel= 2