 Solution

ID: past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2023/1/b/ii/solution

The orthonormal basis makes the squared norm equal to the sum of squared expansion coefficients:

1 = ∣ N ∣^{2} (2 π R) (\frac{1}{4} + \frac{1}{16} + \frac{1}{16}) = ∣ N ∣^{2} \frac{3 π R}{4} .

Choose a positive real wavefunction normalization constant, fixing an irrelevant global phase:

N = \frac{2}{3 π R}, Ψ (0) = \frac{2}{3} ϕ_{0} - \frac{1}{6} (ϕ_{2} + ϕ_{- 2}) .

As a direct check,

\int_{0}^{2 π R} sin^{4} (x / R) d x = 3 π R /4

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