26 | May | 2015 | Physics Town

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 $r_1 + r_2$ .

~~~

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

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

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

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

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

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

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

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

S	M	T	W	T	F	S
					1	2
3	4	5	6	7	8	9
10	11	12	13	14	15	16
17	18	19	20	21	22	23
24	25	26	27	28	29	30
31

Physics Town

continues

Daily Archives: May 26, 2015

Quick Calculation 14.3