2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ What happens to the sum over topologies in this case. In Ginsparg for example, the case where we have both periodic boundary conditions is discuss in great details. There, we need to make a sum over (m,n) which label the two windings around the circle. For the case with other boundary conditions, what happens to this infinite sum over topologies that does not appear anymore ? Thank you very much in advance! $\endgroup$ – Ezareth Mar 23 '20 at 13:36
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Many thanks, and my mistake seems to be that the summation of (n-1/2) was separately done upon n and (-1/2), which produces 1/6. But I cannot quite understand the wrongness of such separate summations. Perhaps, I should go to read the complex analysis more carefully. (By the way, may I point out a trivial typo in your answer where summation should be done over n>0 rather than n>1) $\endgroup$ – Smart Yao Sep 18 '17 at 11:17
  • $\begingroup$ Thanks for spotting the typo. As for the decomposition of the sum, in general this is not possible when regularizing divergent series. You can show that a regularization scheme that is stable ($\sum a_n = a_0 + \sum a_{n+1}$), consistent ($\sum a_n$ gives the expected result when the series is actually convergent) and linear ($\sum (\lambda a_n + b_n) = \lambda \sum a_n + \sum b_n$) is very inefficient (for instance, it can not give a value to $\sum n$). Therefore, one usually drops out the linearity assumption. Indeed, zeta regularization is obviously non-linear. $\endgroup$ – Antoine Sep 18 '17 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.