The free boson conformal field theory has action quadratic in a scalar $X$ and central charge one. Its oscillator partition function is expressed through the <Dedekind eta function>.

Free boson conformal field theory

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

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{title2=$\eta(\tau)$}
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The Dedekind eta function is
$$
\eta(\tau)=q^{1/24}\prod_{n=1}^{\infty}(1-q^n),
\qquad q=e^{2\pi i\tau}.
$$
It packages the oscillator contribution of a chiral free boson.


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