A paradoxical equation in RNS string fermionic part

Question

It is well known for RNS string, $i.e.$, worldsheet supersymmetry formalism, the open string NS sector has worldsheet fermion expansion:

\begin{equation} \psi^{\mu}_{\pm} = \frac{1}{\sqrt 2} \sum_{r \in \mathbb{Z}+\frac{1}{2}} b^{\mu}_r e^{-i r (\tau \pm \sigma)} \end{equation}

The super-Virasoro algebra can be extracted from the energy momentum tensor \begin{equation} T_{++} = (\rm{bosonic ~part}) + \frac{i}{2}\psi_+ \partial_+ \psi_+ \end{equation}

These two equations can be found at Becker, Becker, Schwarz's book, page 121, 123, and is verified by myself. However, when trying to extract the fermionic part of the super-Virasoro algebra, $L_m$, there are two ways to do. One is by direct substituting the mode expansion for $\psi_+$ in to $T_{++}$ and read off the $e^{-i m (\tau+\sigma)}$ coefficent, the other is to use the integration \begin{equation} L_m = \frac{1}{\pi} \int_{-\pi}^{\pi} T_{++} e^{i m \sigma} d\sigma \end{equation}

However, when calculating, we get: \begin{equation} \begin{aligned} \frac{i}{2}\psi_+ \partial_+ \psi_+ & = \frac{i}{4} \sum_{r,s \in \mathbb{Z}+\frac{1}{2}} (-i s)b_r \cdot b_s e^{-i (r+s) (\tau + \sigma)}\\ & = \frac{1}{4}\sum_{m \in \mathbb{Z}}\sum_{r \in \mathbb{Z}+\frac{1}{2}}(m-r) b_r \cdot b_{m-r}e^{-i m (\tau+\sigma)}\\ & = \frac{1}{4}\sum_{m \in \mathbb{Z}}\sum_{r \in \mathbb{Z}+\frac{1}{2}}(m+r) b_{-r} \cdot b_{m+r}e^{-i m (\tau+\sigma)} \end{aligned} \end{equation}

and one then put normal ordering. Here $\partial_+ = 1/2(\partial_{\tau}+\partial_{\sigma})$. In the book, page 126, the coefficent is actually $(m+2r)$. I also checked GSW's book, as well as Polchinski's book, which indicates I am wrong at some point. But after long time looking for it I do not know why.

It is a very stupid and technical question, but I have been stuck to it for some time. So it would be great that someone may help solve it.

This post imported from StackExchange Physics at 2014-10-23 07:29 (UTC), posted by SE-user Kevin Ye

Answer 1

The normal ordering makes the expression of $b_{-r}\cdot b_{r+m}$ symmetric around the point $r=-\frac{m}{2}$, and these symmetric parts differ by a sign. This sign difference due to normal ordering makes $m/2$ redundancy.

This post imported from StackExchange Physics at 2014-10-23 07:29 (UTC), posted by SE-user Kevin Ye