# Strange factor multiplying the fermionic part in the NS mass-squared operator?

+ 1 like - 0 dislike
15 views

In the Neveu-Schwarz sector, the worldsheet fermions can be expanded as

$$\psi^I(\tau,\sigma) \sim \sum\limits_{r\in Z+1/2}b_r^Ie^{-ir(\tau-\sigma)}$$

and the total mass squared operator can then be written as

$$M^2 = \frac{1}{\alpha'}\left( \frac{1}{2} \sum\limits_{p\neq 0} \alpha_{-p}^I\alpha_p^I + \frac{1}{2}\sum\limits_{r\in Z+1/2} r \, b_{-r}^I b_r^I \right)$$

The first sum gives the contribution of the bosons, the second one the contributions of the fermions.

Why are the summands in in the fermionic part multiplied by $r$, how does this factor come in mathematically? Does this have something to do with the Pauli exclusion principle?

The same thing issue appears with the fermionic part mass operator in the Ramond sector, where I dont understand it either ...

retagged Mar 25, 2014
Funny, the proof of this has not even been given in BBS, maybe I should check my trustable lecture notes. The "lecture" is the trustiest/.

This post imported from StackExchange Mathematics at 2014-03-09 15:47 (UCT), posted by SE-user dimensio1n0
@DImension10AbhimanyuPS if you find the answer, you could post it here :-). When trying to look at the lecture notes, the nasty firewall we have at work (that should not exist as two young bright colleagues of Lumo nicely proved ...) intervenes, so I'll look at it at home.

This post imported from StackExchange Mathematics at 2014-03-09 15:47 (UCT), posted by SE-user Dilaton
Unfortunately, all three lecture notes betrayed me, as did BBS and Mc Mohan : ( However, I think I found the answer myself, and will be posting it here soon.

This post imported from StackExchange Mathematics at 2014-03-09 15:47 (UCT), posted by SE-user dimensio1n0

If one solves the field equations for the bosonic field, with the Newmann/Dirchilet/Closed String Boundary conditions, one can see that the mode expansion is something like: $$X^\mu=...+i\sqrt{2\alpha'} \sum_{n\neq0 }^{ } \frac{\alpha^\mu}{n}\exp\left(in\sigma^0\right)\cos\left(n\sigma^1\right)$$ On the other hand, the fermionic field mode expansion goes like: $$\psi^\mu_\pm = \frac1{\sqrt2}\sum_{n\in\mathbb Z \ \mathrm{or} \ \mathbb{Z}+\frac12 }b_r^\mu \exp\left(-ir\sigma^\pm\right)$$ Notice that there is a missing factor of $\frac1r$ in the second equation. $[N,b_{-r}^\mu]=rb_{-r}^\mu$ and the conclusion follows.
 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:p$\hbar$ysi$\varnothing$sOverflowThen 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.