# Quick Calculation 14.3

A First Course in String Theory

To test the $N^\perp$ includes the fermionic contribution to the number, show that the eigenvalue of $N^\perp$ on $b_{-r_1}^I b_{-r_2}^J | NS \rangle$, with $r_1, r_2 > 0$, is $r_1 + r_2$.

~~~

$N^\perp b_{-r_1}^I b_{-r_2}^J |NS \rangle$

$= \left[ \sum_{I'} \left( \sum_{p=1}^\infty \alpha_{-p}^I \alpha_p^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r b_{-r}^I b_r^I \right) \right] b_{-r_1}^I b_{-r_2}^J |NS \rangle$

$= \sum_{I'} \left( \sum_{p=1}^\infty \alpha_{-p}^{I'} \alpha_p^{I'} + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r b_{-r}^{I'} b_r^{I'} \right)$ $|n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

$= \sum_{I'} \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r b_{-r}^{I'} b_r^{I'} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

$= \sum_{I'} [ r_1 b_{-r_1}^{I'} b_{r_1}^{I'} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle + r_2 b_{-r_2}^{I'} b_{r_2}^{I'} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle]$

$= r_1 b_{-r_1}^{I} b_{r_1}^{I} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle + r_2 b_{-r_2}^{I} b_{r_2}^{I} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

$+ r_1 b_{-r_1}^{J} b_{r_1}^{J} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle + r_2 b_{-r_2}^{J} b_{r_2}^{J} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

$= r_1 b_{-r_1}^{I} b_{r_1}^{I} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle + 0 + 0 + r_2 b_{-r_2}^{J} b_{r_2}^{J} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

— Me@2015-05-25 12:52:23 AM