= Solution
Put $\mathcal T=\beta\hbar$ and $x=\omega\mathcal T$. With equal endpoints $q$, the Euclidean classical path and its action are
$$
q_{\rm cl}(\tau_E)=q\frac{\cosh[\omega(\tau_E-\mathcal T/2)]}{\cosh(x/2)},\qquad S_{\rm cl}=m\omega q^2\tanh(x/2).
$$
Write $q(\tau_E)=q_{\rm cl}+\eta$, where $\eta(0)=\eta(\mathcal T)=0$. Sine modes have fluctuation eigenvalues $(n\pi/\mathcal T)^2+\omega^2$. Relative to the free-particle determinant, their Gaussian integral contributes
$$
\prod_{n=1}^\infty\left[1+\left(\frac{x}{n\pi}\right)^2\right]^{-1/2}=\sqrt{\frac{x}{\sinh x}}.
$$
Multiplying the free prefactor $\sqrt{m/(2\pi\hbar\mathcal T)}$ gives the <Euclidean oscillator diagonal kernel>
$$
\boxed{K_E(q,q;\beta\hbar)=\sqrt{\frac{m\omega}{2\pi\hbar\sinh x}}\exp\!\left[-\frac{m\omega q^2}{\hbar}\tanh(x/2)\right].}
$$
Integrating this over the common endpoint is precisely the trace. All fluctuations here are Dirichlet fluctuations about a fixed-endpoint path; periodicity is imposed on the full trace after endpoint integration.
Back to article page
