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

For the Hamiltonian $p^2/(2m)+V(q)$, the real-time configuration-space path integral iswith its measure defined by time slicing. Wick rotation gives the Euclidean path integral and the thermal trace.

For the infinite square well on $(0,L)$, the Dirichlet boundary conditions select the normalized energy eigenstatesThere is no $n=0$ state: the corresponding sine is identically zero. Taking the trace in this orthonormal basis gives the canonical partition function

Use unfolding a reflected interval trajectory: reflect the interval across each wall so that a bouncing trajectory becomes a free trajectory in an image interval. A return path starting at $q$ ends at $q+2rL$ after an even number of reflections, or at $-q+2rL$ after an odd number. Each hard-wall reflection contributes phase $-1$, giving the positive direct images and negative reflected images of the Dirichlet heat kernel on an interval.

The method of images therefore givesHere $\tau$ has inverse-energy units, since $\beta =1/(k_{B}T)$. In physical imaginary time $u=\hslash \tau$, the same dimensionless action is $\mathcal{A}=S_E/\hslash$, with $S_E={\int }_0^{\hslash \beta }(m/2)(dq/du{)}^{2}du$. Thus the question's convention $S=(m/(2\hslash )){\int }_0^{\beta }({\partial }_{\tau }q{)}^{2}d\tau$ has exactly the weight $e^{-S/\hslash }$.

Analytically continuing the free-particle propagator to imaginary time gives the Gaussian heat kernelThe two image endpoints have displacements $2rL$ and $2rL-2q$ from the starting point. Inserting their kernels into the thermal trace of an interval image kernel givesThe prefactor comes from the normalization of the free-particle propagator; it cannot be dropped from a thermal trace.

Set $a=2m{L}^2/({\hslash }^2\beta )$ and $C=\sqrt{m/(2\pi {\hslash }^2\beta )}$. For the reflected images, the intervals $rL-q$, $0\le q\le L$, tile the real line once. A Gaussian integral givesConsequently $Z=CL{\sum }_r{e}^{-a{r}^2}-1/2$. The Poisson summation formula applied to the Gaussian yieldsThe zero dual mode cancels the reflected contribution, while the positive and negative modes pair:This proves equality with the energy eigenstate calculation. The image expansion converges rapidly at small $\beta$; the spectral expansion converges rapidly at large $\beta$.