 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2024/paper-1/a1/a/iv/solution

n, m

, use the product-to-sum identity in the inner product:

⟨ Ψ_{m} ∣ Ψ_{n} ⟩ = \frac{2}{a} \int_{0}^{a} sin \frac{mπ x}{a} sin \frac{nπ x}{a} d x = \frac{1}{a} \int_{0}^{a} [cos \frac{( m - n ) π x}{a} - cos \frac{( m + n ) π x}{a}] d x = [\frac{sin (( m - n ) π x / a )}{( m - n ) π} - \frac{sin (( m + n ) π x / a )}{( m + n ) π}]_{0}^{a} = 0.

Every endpoint sine vanishes. For

m = n

, the normalization integral is one; the expression with a denominator

m - n

should not be substituted directly. Thus the normalized functions form an orthonormal set,

⟨ Ψ_{m} ∣ Ψ_{n} ⟩ = δ_{mn}

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