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

+ 1 like - 0 dislike

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 ...

asked Oct 27, 2013 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]
retagged Mar 25, 2014 by dimension10
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

1 Answer

+ 1 like - 0 dislike
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.
answered Oct 28, 2013 by dimension10 (1,985 points) [ revision history ]
Gosh, I'll have to check if the factor 1/r and maybe a factor r in the commutator is missing in my Zweibach book too or if it is just missing in my question ... Your answer will then solve the issue, thanks :-). In fact, thnking about it I should have seen that it is missing in the fermion expansion myself, darn ...!

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

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