 Solution

Because $S[x_0]=0$ at $x=0$, the factor $Z$ cancels from the ratio. Write $z=\beta \hslash \omega$ and $q=\sqrt{m\omega /\hslash }x$. The exact answer isAt high temperature, $z\ll 1$. Expanding the hyperbolic tangent givesThe displayed leading expression is the ratio of classical Boltzmann factors for potential energy $V(x)=m{\omega }^2{x}^2/2$ and $V(0)=0$. Indeed the classical canonical ensemble givesThe momentum Gaussian integral is independent of $x$ and cancels in the ratio. This is the classical phase-space limit of the thermal kernel. Here $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant; the associated classical position variance is $1/(\beta m{\omega }^2)$. At fixed $x$, $R$ ultimately tends to one as temperature tends to infinity, while its width grows.

At low temperature, $z\gg 1$. Now $\tanh (z/2)=1-2e^{-z}+O(e^{-2z})$, so at fixed $q$Unlike the classical high-temperature expression, this limiting width remains nonzero at zero temperature, reflecting the ground state of the quantum harmonic oscillator.