 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ib/paper-1/15a/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

The Parseval identity for the orthonormal stationary-state expansion is

\sum_{n \geq 1} ∣ c_{n} ∣^{2} = ∥ ψ ∥^{2} = 1

. Splitting the amplitudes into even and odd indices gives

1 = \frac{3}{π ^{2}} \sum_{p = 1}^{\infty} \frac{1}{p ^{2}} + \frac{12}{π ^{2}} \sum_{p = 1}^{\infty} \frac{1}{( 2 p - 1 ) ^{2}} [1 + \frac{2 ( - 1 ) ^{p}}{( 2 p - 1 ) π}]^{2} .

Multiply by

π^{2} /3

to obtain the requested identity. Its integer constants are

A = 1, B = 4, C = 2.

The series converges absolutely, since both parity contributions are

O (p^{- 2})

. This derivation uses the normalization of the prepared state and completeness of the well stationary states, rather than a separate summation trick.

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