= Solution
Insert the complete <orthonormal basis> of <energy eigenstates> on either side of the <Boltzmann factor>. The <spectral theorem> gives
$$
e^{-\beta H}=\sum_{n=0}^{\infty}e^{-\beta\hbar\omega(n+1/2)}|n\rangle\langle n|,
\qquad
\boxed{\langle x|e^{-\beta H}|x\rangle
=\sum_{n=0}^{\infty}e^{-\beta\hbar\omega(n+1/2)}|\varphi_n(x)|^2.}
$$
At low <temperature>, the gap $\hbar\omega$ suppresses excited <energy eigenstates>. The <harmonic oscillator eigenfunction> of the <ground state> is
$$
\varphi_0(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}
e^{-m\omega x^2/(2\hbar)}.
$$
Therefore
$$
\langle x|e^{-\beta H}|x\rangle
\sim e^{-\beta\hbar\omega/2}|\varphi_0(x)|^2,
\qquad
\boxed{R\longrightarrow\frac{|\varphi_0(x)|^2}{|\varphi_0(0)|^2}
=e^{-m\omega x^2/\hbar}.}
$$
The common <zero-point energy> factor cancels. Dividing the kernel by its <trace> would give the position <probability density> of the <canonical ensemble>; at zero <temperature> this tends to $|\varphi_0(x)|^2$, whose <variance> is $\hbar/(2m\omega)$. Thus zero <temperature> projects onto the <ground state> and retains its quantum spread, rather than concentrating at the minimum of the classical <potential energy>.
There is also a check on the first correction. With $q=\sqrt{m\omega/\hbar}\,x$, the first excited <harmonic oscillator eigenfunction> satisfies $|\varphi_1(x)|^2=2q^2|\varphi_0(x)|^2$ and $\varphi_1(0)=0$. Hence the <spectral theorem> gives $R=e^{-q^2}[1+2q^2e^{-z}+O(e^{-2z})]$, in agreement with part (c).
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