Path integral of $Z_2$-orbifold free boson on torus

Question

The partition function of $Z_2$-orbifold free boson on 2-torus (parametrized by $\tau\equiv\omega_2/\omega_1$) is derived by operator formalism. I want to consider it in the path-integral approach: \begin{eqnarray} Z=\sum_{\nu,u}\prod_{m,n}\sqrt{\frac{\pi}{-\lambda_{m,n}^{(\nu,u)}}} \end{eqnarray} where $\nu$ and $u$ taking values among $0$ and $1/2$ define various boundary conditions \begin{eqnarray} \phi(z+\omega_1)=\exp(i2\pi\nu)\phi(z), \\ \phi(z+\omega_2)=\exp(i2\pi u)\phi(z). \end{eqnarray} $\lambda_{m,n}^{(\nu,u)}$'s are the eigenvalues of Laplacian $\triangledown_\mu\triangledown^\mu$ with \begin{eqnarray} \lambda_{m,n}^{(\nu,u)}=-\frac{4\pi^2[(m+\nu)-(n+u)\tau][(m+\nu)-(n+u)\bar\tau]}{L^2(\text{Im}\tau)^2}. \end{eqnarray} For simplicity, we can consider the sector in which $\nu=0$ while $u=1/2$ since in this section there are no zero modes to worry about and $m,n$ can take any integer. Then \begin{eqnarray} Z_{0,1/2}&=&\prod_{m\in\mathbb{Z},n\in\mathbb{Z}}\sqrt{\frac{\pi}{-\lambda_{m,n}^{(0,1/2)}}}\\ &=&\sqrt{\frac{1}{\prod_{m\in\mathbb{Z},n\in\mathbb{Z}}[m-(n+1/2)\tau][m-(n+1/2)\bar\tau]}} \end{eqnarray} where $\prod_{n=-\infty}^\infty$a=1 is made use of. To calculate the denominator, one can define $q=\exp(i2\pi\tau)$ and then \begin{eqnarray} &&\prod_{m\in\mathbb{Z},n\in\mathbb{Z}}[m-(n+1/2)\tau][m-(n+1/2)\bar\tau]\\ &=&\prod_{n}\left\{\left[q^{-(n+1/2)/2}-q^{(n+1/2)/2}\right]\times\text{h.c.}\right\}\\ &=&\prod_{n\geq0}\left\{\left[q^{-(n+1/2)/2}-q^{(n+1/2)/2}\right]\times\text{h.c.}\right\}\prod_{n>0}\left\{\left[q^{-(n-1/2)/2}-q^{(n-1/2)/2}\right]\times\text{h.c.}\right\}\\ &=&\prod_{n>0}\left\{\left[q^{-(n-1/2)/2}-q^{(n-1/2)/2}\right]\times\text{h.c.}\right\}^2\\ &=&(q\bar{q})^{-1/6}\prod_{n=1}^\infty(1-q^{n-1/2})^2(1-\bar{q}^{n-1/2})^2 \end{eqnarray} which means \begin{eqnarray} Z_{0,1/2}=\frac{(q\bar{q})^{1/12}}{\prod_{n=1}^\infty(1-q^{n-1/2})(1-\bar{q}^{n-1/2})}. \end{eqnarray} However, in the conventional introduction to conformal field theory, such as Eq.~(8.23) of ``Applied conformal field theory'' by P. Ginsparg, there is a $\tau$-dependence phase difference: \begin{eqnarray} \tilde{Z}_{0,1/2}=\frac{(q\bar{q})^{1/48}}{\prod_{n=1}^\infty(1-q^{n-1/2})(1-\bar{q}^{n-1/2})}. \end{eqnarray} I really cannot understand the existence of this little but essential difference. I notice that the phase difference is $1/16$ which might have its origin in the zero mode change of Virasoro generator $L_0$ when twisting the boundary condition, but I cannot let it make sense quantatively.

Answer 1

You have to use zeta regularization to compute the sum $$\sum\limits_{n>0} \left( n - \frac{1}{2}\right) \, . $$ In other words, evaluate the regularized sum $$\sum\limits_{n>0} \left( n - \frac{1}{2}\right)^{-s} = (2^s -1) \zeta (s) \rightarrow \frac{1}{24} $$ where the limit $s \rightarrow -1$ is taken.

After taking the square root, you obtain $(q \bar{q})^{1/48}$ as wanted.