 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-81/3/d/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

a = 2 m L^{2} / (ℏ^{2} β)

and

C = m / (2 π ℏ^{2} β)

. For the reflected images, the intervals

r L - q

0 \leq q \leq L

C \sum_{r} \int_{0}^{L} e^{- 2 m (r L - q)^{2} / (ℏ^{2} β)} d q = C \int_{- \infty}^{\infty} e^{- 2 m x^{2} / (ℏ^{2} β)} d x = \frac{1}{2} .

Consequently

Z = C L \sum_{r} e^{- a r^{2}} - 1/2

. The Poisson summation formula applied to the Gaussian yields

\sum_{r \in Z} e^{- a r^{2}} = \frac{π}{a} \sum_{n \in Z} e^{- π^{2} n^{2} / a}, C L \frac{π}{a} = \frac{1}{2} .

The zero dual mode cancels the reflected contribution, while the positive and negative modes pair:

Z = \frac{1}{2} n \in Z \sum e^{- β ℏ^{2} π^{2} n^{2} / (2 m L^{2})} - \frac{1}{2} = n = 1 \sum \infty e^{- β ℏ^{2} π^{2} n^{2} / (2 m L^{2})} .

This proves equality with the energy eigenstate calculation. The image expansion converges rapidly at small

β

; the spectral expansion converges rapidly at large

β

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