Put $\tau=\tau_1+i\tau_2$ and $q=e^{2\pi i\tau}$. The torus partition function is the <trace>
$$
Z(\tau,\bar\tau)=\operatorname{Tr}
e^{-2\pi\tau_2H+2\pi i\tau_1P}.
$$
The continuum momentum states have density $V/(2\pi)$, so their contribution is
$$
\frac V{2\pi}\int_{-\infty}^{\infty}dp\,e^{-\pi\tau_2p^2}
=\frac V{2\pi\sqrt{\tau_2}}.
$$
Each left-moving oscillator contributes $\sum_{N\geq0}q^{nN}=(1-q^n)^{-1}$, with the complex conjugate for the right mover. The zero-point factor combines these products into the <Dedekind eta function>
$$
\eta(\tau)=q^{1/24}\prod_{n=1}^{\infty}(1-q^n).
$$
Therefore the <torus partition function of a free boson> is
$$
\boxed{Z(\tau,\bar\tau)=\frac V{2\pi\sqrt{\operatorname{Im}\tau}}
|\eta(\tau)|^{-2}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 306 / 2 / iii / Solution

For a noncompact free boson with target volume $V$, the torus partition function is $Z=V[2\pi\sqrt{\operatorname{Im}\tau}]^{-1}|\eta(\tau)|^{-2}$ in the normalization where the zero-mode Hamiltonian is $p^2/2$.


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