Solution (source code)

= Solution

For the <quantum harmonic oscillator>, set $\mu=0$ and $\Omega=\hbar\omega$. The normal symbol is $H_N(\bar\psi,\psi)=\Omega(\bar\psi\psi+1/2)$, so the thermal <coherent-state time slicing> gives
$$
Z=\int_{\mathrm{periodic}}\mathcal D(\bar\psi,\psi)\exp\left[-\int_0^\beta\big(\bar\psi\partial_\tau\psi+\Omega\bar\psi\psi+\Omega/2\big)d\tau\right].
$$
The connection with a <phase-space path integral> uses the canonical real coordinates
$$
\psi=\frac{\sqrt{m\omega}\,q+ip/\sqrt{m\omega}}{\sqrt{2\hbar}},\qquad \bar\psi=\frac{\sqrt{m\omega}\,q-ip/\sqrt{m\omega}}{\sqrt{2\hbar}}.
$$
Their derivative term satisfies $\bar\psi\dot\psi=-ip\dot q/\hbar$ up to total derivatives that vanish for periodic paths. The <normal and Weyl symbols of a harmonic oscillator> must be distinguished: in midpoint phase-space time slicing, the <Weyl ordering> symbol of $a^\dagger a$ is $\bar\psi\psi-1/2$. Thus the appropriate midpoint Hamiltonian is $H_W=p^2/(2m)+m\omega^2q^2/2$, with no additional constant. The change from the adjacent-label normal prescription to the midpoint prescription includes this ordering correction.

In physical imaginary time $u=\hbar\tau$, the resulting <phase-space path integral> is
$$
Z=\int_{\mathrm{periodic}}\mathcal Dq\,\mathcal Dp\,\exp\left[-\frac1\hbar\int_0^{\hbar\beta}\left(-ip\frac{dq}{du}+\frac{p^2}{2m}+\frac{m\omega^2q^2}{2}\right)du\right].
$$
<Gaussian momentum integration in a phase-space path integral> produces the usual oscillator <configuration-space path integral>. An exact check follows directly from the coherent kernel $\langle\psi'|e^{-\beta\Omega a^\dagger a}|\psi\rangle=\exp(e^{-\beta\Omega}\bar\psi'\psi)$:
$$
\boxed{Z=e^{-\beta\Omega/2}\int_{\mathbb C}\frac{d^2\psi}{\pi}e^{-(1-e^{-\beta\Omega})|\psi|^2}=\frac{e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}}=\frac1{2\sinh(\beta\hbar\omega/2)}}.
$$
This is the <thermal partition function of a quantum harmonic oscillator>. Keeping the $\Omega/2$ normal-ordering constant a second time after switching to the Weyl symbol would double count the zero-point energy.