224
3 Quantum Mechanics – II
where F(r ) = r f (r )
L z ψ 1 = −i
∂
∂ϕ
(cos ϕ + i sin ϕ) sin θ F(r )
= −i(− sin ϕ + i cos ϕ) sin θ F(r )
= (cos ϕ + i sin ϕ) sin θ F(r )
= ψ 1
Thus ψ 1 is the eigen state and the eigen value is
ψ 2 = z f (r ) = r cos θ f (r )
L z ψ 2 = −i
∂
∂ϕ
(r cos θ f (r )) = 0
The eigen value is zero.
ψ 3 = (x − iy) f (r ) = r sin θ(cos ϕ − i sin ϕ) f (r )
= (cos ϕ − i sin ϕ) sin θ F(r )
L z ψ 3 = −i
∂
∂ϕ
(cos ϕ − i sin ϕ) sin θ F(r )
= −i(− sin ϕ − i cos ϕ) sin θ F(r )
= i (sin ϕ + i cos ϕ) sin θ F(r )
= −(cos ϕ − i sin ϕ) sin θ F(r ) = −ψ 3
Thus ψ 3 is an eigen state and the eigen value is −.
3.89 (a) L z = −i
∂
∂ϕ
L z ψ = −i
∂
∂ϕ
Af (r ) sin θ cos θ e
iϕ
= (i) (−i A f (r ) sin θ cos θ e
iϕ )
= A f (r ) sin θ cos θ e
iϕ
= ψ
Therefore, the z-component of the angular momentum is .
(b) L
2
= −
2
∂
2
∂θ 2 + cot θ
∂
∂θ
+
1
sin
2 θ
∂
2
∂ϕ 2
Expressions for L z and L
2 are derived in Problems 3.80 and 3.83.
L
2 ψ = −
2
∂
2
∂θ 2 + cot θ
∂
∂θ
+
1
sin
2 θ
·
∂
2
∂ϕ 2
Af (r ) sin θ cos θ e
iϕ
= −
2 Af (r )e
iϕ
−4 sin θ cos θ + cot θ (cos
2 θ − sin
2 θ) −
sin θ cos θ
sin
2 θ
= 6
2 Af (r ) sin θ cos θ e
iϕ
= 6
2 ψ
Thus L
2
= 6
2
But L
2
= l(l + 1). Therefore l = 2