Past exam of the advanced quantum theory: path integrals and many-particle physics course of the University of Oxford / 2024 / 1 / b / Solution

After the shift $x(\tau )=x_0(\tau )+y(\tau )$, the endpoint values of $y$ are both zero. Expanding the Euclidean action givesBy integration by parts, the cross term isThe first term vanishes by the homogeneous Dirichlet boundary conditions, and the second by the Euler-Lagrange equation. The translation has unit Jacobian determinant in the time-sliced configuration-space path integral, so $\mathcal{D}x=\mathcal{D}y$. Consequently classical-path factorization of a quadratic path integral is exact here:All dependence on the common endpoint $x$ lies inThe endpoint domain and the integrand defining $Z$ are the same for every $x$. This $Z$ is a zero-endpoint Gaussian path integral; the canonical partition function instead integrates the entire diagonal kernel over $x$. The final sentence of the printed part omits $/\hslash$ from its exponential: the physical Boltzmann factor has the dimensionless exponent $-S/\hslash$.

An optional normalization check uses the Dirichlet oscillator fluctuation determinant. Its eigenvalues are $m[(n\pi /\mathcal{T}{)}^2+{\omega }^2]$, $n\ge 1$. Relative to the free-particle Gaussian path integral, the hyperbolic-sine infinite product givesThus $Z=\sqrt{m\omega /[2\pi \hslash \sinh (\beta \hslash \omega )]}$. Integrating the resulting diagonal thermal kernel of a quantum harmonic oscillator by a Gaussian integral yields $\mathrm{Tr}{e}^{-\beta H}=[2\sinh (\beta \hslash \omega /2){]}^{-1}$, including the zero-point energy.