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264 7 QUANTUM THEORY
as the integrand can be written as the product of a function of θ with a
function of ϕ, the integral can be separated
= NaNb ∫
2π
0
sin ϕ cos ϕ dϕ∫
π
0
sin3 θ dθ
= (NaNb/2) × sin2 ϕ∣
2π
0 × ∫
π
0
sin3 θ dθ
= (NaNb/2) × [sin2(2π) − sin2(0)] × ∫
π
0
sin3 θ dθ
= 0
Hence, ψa and ψb are orthogonal.
(d) �e wavefunctions are normalized if
∫
2π
ϕ=0
∫
π
θ=0
ψ∗aψa sin θ dθ dϕ = 1 and ∫
2π
ϕ=0
∫
π
θ=0
ψ∗bψb sin θ dθ dϕ = 1
Note that both ψa and ψb are real. For ψa this integral is
I =∫
2π
ϕ=0
∫
π
θ=0
(Na sin θ cos ϕ)2 sin θ dθ dϕ = N2a ∫
2π
ϕ=0
∫
π
θ=0
sin3 θ cos2 ϕ dθ dϕ
�is integral can be separated
I = N2a ∫
2π
0
cos2 ϕ dϕ∫
π
0
sin3 θ dθ = N2a ∫
2π
0
1 − sin2 ϕ dϕ∫
π
0
sin3 θ dθ
�e integral in ϕ can be evaluated using Integral T.3 and the integral in θ
using Integral T.2
I = N2a[2π − 2π/2 + (1/4) sin 2(2π)](1/3)[2 − (sin2 π + 2) cos π]
= N2a(π)(4/3)
�e normalizing factor is therefore Na =
√
3/4π .
For ψb ,
I =∫
2π
ϕ=0
∫
π
θ=0
(Nb sin θ sin ϕ)2 sin θ dθ dϕ = N2b ∫
2π
ϕ=0
∫
π
θ=0
sin3 θ sin2 ϕ dθ dϕ
�is integral can be separated
I = N2b ∫
2π
0
sin2 ϕ dϕ∫
π
0
sin3 θ dθ
�e integral in ϕ is evaluated using Integral T.3 and the integral in θ using
Integral T.2
I = N2b[2π + 2π/2 − (1/4) sin 2(2π)](1/3)[2 − (sin2 π + 2) cos π]
= N2b(π)(4/3)
�e normalizing factor is therefore Nb =
√
3/4π .