 Solution

Put $\mathcal{T}=\beta \hslash >0$, the interval of imaginary time, and assume $m,\omega >0$. For a variation $x_0+ϵ\eta$ with $\eta (0)=\eta (\mathcal{T})=0$, the first variation of the Euclidean action isThe boundary term vanishes. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationIts sign differs from the real-time harmonic oscillator equation because the Euclidean action contains positive kinetic energy and positive potential energy. The general solution is $A\cosh (\omega \tau )+B\sinh (\omega \tau )$. Applying the two endpoint conditions gives $A=x$ and $B=-x\tanh (\omega \mathcal{T}/2)$. Equivalently, the equal-endpoint Euclidean oscillator path isThis form displays reflection about the midpoint and satisfies both endpoints immediately. On this path,Thus the Euclidean action reduces to a boundary term. Since ${\dot{x}}_0(\mathcal{T})=\omega x\tanh (\omega \mathcal{T}/2)$ and ${\dot{x}}_0(0)=-\omega x\tanh (\omega \mathcal{T}/2)$,The positive quadratic form $m{\int }_0^{\mathcal{T}}({\dot{\eta }}^2+{\omega }^2{\eta }^2)d\tau$ is the second variation, so this stationary path is the unique minimum, rather than merely an arbitrary stationary point.