# Quick Calculation 14.8

A First Course in String Theory

What sector(s) can be combined with a left-moving NS- to form a consistent closed string sector?

~~~

— This answer is my guess. —

What is the meaning of “left-moving NS”?

p.322 “Conventionally, the first input in $(\cdot, \cdot)$ is the left-moving sector and the second input is the right-moving sector.”

What is the meaning of “+” in “NS+”?

p.321 “… we truncate the NS sector to the set of states with $(-1)^F = +1$. The resulting states comprise the so-called NS+ sector.”

“The NS+ sector contains the massless states and throw away the tachyonic states.”

What is the meaning of “consistent” closed string sector?

p.322 “… guarantees that the left and right sectors give identical contributions to the mass-squared:” $\alpha' M_L^2 = \alpha' M_R^2$

If NS+ is the left sector, the corresponding candidate right sectors are NS+, R+, R-. $M$ in NS-sector, Equation (14.37): $M_{NS}^2 = \frac{1}{\alpha'} \left( - \frac{1}{2} + N^\perp \right)$ $M$ in R-sector, Equation (14.53): $M_{Ramond}^2 = \frac{1}{\alpha'} \sum_{n \ge 1} \left( \alpha_{-n}^I \alpha_n^I + n d_{-n}^I d_n^1 \right) = \frac{1}{\alpha'} N^\perp$

Equation (14.78): $\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2$

NS+ equations of (14.38): $\alpha'M^2=0,~~N^\perp = \frac{1}{2}:~~~~b_{-1/2}^I |NS \rangle \otimes |p^+, \vec p_T \rangle$, $\alpha'M^2=1,~~N^\perp = \frac{3}{2}:~~~~\{ ... \} |NS \rangle \otimes |p^+, \vec p_T \rangle$, $...$

NS- equations of (14.38): $\alpha'M^2=-\frac{1}{2},~~N^\perp = 0:~~~~\{ ... \} |NS \rangle \otimes |p^+, \vec p_T \rangle$, $\alpha'M^2=\frac{1}{2},~~N^\perp = 1:~~~~\{ ... \} |NS \rangle \otimes |p^+, \vec p_T \rangle$, $...$

Mass levels of R+ and R- (Equations 14.54): $\alpha'M^2=0,~~N^\perp = 0:~~~~|R_a \rangle~~||~~|R_{\bar a} \rangle$ $\alpha'M^2=1,~~N^\perp = 1:~~~~...$ $...$

There are no mass levels in NS+, R+, or R- that can match those in NS-. So NS- can be paired only with NS-:

(NS-, NS-)

— This answer is my guess. —

— Me@2015-07-05 11:09:32 PM