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I am trying to derive the quantum form of the Virasoro constraints. $$ L_{m} = \frac{1}{2} \sum_{n} :\alpha_{m-n}.\alpha_{n}: $$ :...: meaans normal ordering. Using the common commutator between modes of string when $ m=0 $ $$ [\alpha^{\mu}_{n}, \alpha^{\nu}_{-n}]= n \eta^{\mu\nu} $$ we easily get: $$ L_{0}= \frac{1}{2}\sum_{n=-\infty}^{\infty} \alpha_{-n}.\alpha_{n}= \frac{1}{2} \alpha_{0}^2 + \sum_{n=1}^{\infty} \alpha_{-n}.\alpha_{n}\\ +\frac{D}{2} \sum_{n=1}^{\infty} n. $$ My problem is here when calculating the last term. We know that it is infinite, forgetting it for a moment because text says we can use regularization to calculate its finite value. The simple calculation comes first

$$ \frac{d}{da}\sum_n e^{-na} ~=~\sum_n ne^{-na}~=~ \frac{d}{da} \frac{1}{1-e^{-a}} ~=~ \frac{e^{-a}}{(1-e^{-a})^{2}} ~=~ \frac{1}{a^2}-\frac{1}{12}.$$

First, I could not understand $(\frac{1}{a^2}-\frac{1}{12})$ part. Second, the text concludes that: $\frac{D}{2} \sum_{n=1}^{\infty} n= -\frac{D}{24} $ where I could not follow. So my clear question is "how the regularization works"? It is said that $ a= 2πε /\ell $.

References:

  1. David MacMahon, String theory demystified, p. 79-80.
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There are several typos in MacMahon's formula on the bottom of page 79. It should read

$$ \color{red}{-}~\frac{d}{da}\sum_{n=0}^{\infty} e^{-na} ~=~\sum_{n=0}^{\infty}ne^{-na}~=~ \color{red}{-}~\frac{d}{da} \frac{1}{1-e^{-a}} ~=~ \frac{e^{-a}}{(1-e^{-a})^{2}} ~=~ \frac{1}{a^2}-\frac{1}{12}\color{red}{+{\cal O}(a^2)}.$$ The main point is that the regular terms ${\cal O}(a^2)$ vanish as we remove the regularization $a\to 0^+$, while the singular term $\frac{1}{a^2}$ is cancelled by local counterterms. See e.g. this Math.SE post, this & this Phys.SE posts, and links therein.

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  • $\begingroup$ If you just add a minus at the end of the first line every thing will be right. it was very helpful and I got it. $\endgroup$ – Ali Mar 7 at 23:51

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