 Solution

For unnormalized coherent states, $|z⟩=e^{z{a}^{†}}|0⟩$, $⟨w|z⟩=e^{\bar{w}z}$ and $1=\int d^{2}z{e}^{-|z{|}^2}|z⟩⟨z|/\pi$. Insert this identity between short imaginary-time evolutions. The overlaps generate $\bar{\varphi }{\partial }_{\tau }\varphi$, while the normal-ordered symbol of $H=\hslash \omega (a^{†}a+1/2)$ givesHere $\tau ={\tau }_E/\hslash$ runs from zero to $\beta$; the exponent is dimensionless. The bosonic trace makes the complex fields periodic, not antiperiodic.

The printed expression omits the explicit vacuum factor. It represents the full oscillator only if a symmetric/Weyl determinant prescription implicitly includes that factor. With the ordinary normal-ordered time-sliced coherent-state measure, it instead represents $H=\hslash \omega a^{†}a$. The continuum formula alone does not specify the energy-zero convention.