A paradoxical equation in RNS string fermionic part

+ 3 like - 0 dislike
644 views

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

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

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

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 $$L_m = \frac{1}{\pi} \int_{-\pi}^{\pi} T_{++} e^{i m \sigma} d\sigma$$

However, when calculating, we get: \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}

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

edited Oct 23, 2014

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.
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:$\varnothing\hbar$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.