Solution

Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 81 / 3 / c / Solution

{title2=$Z=\int_0^LK_D(q,q;\beta)dq$}

For a particle with mass $m$ on $(0,L)$, put $K_0(x;\beta)=\sqrt{m/(2\pi\hbar^2\beta)}e^{-mx^2/(2\hbar^2\beta)}$. The <method of images> gives $K_D(q_f,q_i;\beta)=\sum_r[K_0(q_f-q_i+2rL;\beta)-K_0(q_f+q_i+2rL;\beta)]$. The diagonal integral is
$$
Z=L\sqrt{\frac m{2\pi\hbar^2\beta}}\sum_{r\in\mathbb Z}e^{-2mL^2r^2/(\hbar^2\beta)}-\frac12.
$$
The reflected intervals tile the real line, giving the subtraction $1/2$. The <Poisson summation formula> converts this to $\sum_{n\ge1}e^{-\beta\hbar^2\pi^2n^2/(2mL^2)}$, proving equality between the image and <energy eigenstate> calculations.

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 Incoming links: Thermal trace of an interval image kernel