Problem 14.4a

Closed string degeneracies.

For closed string states the left-moving and right-moving excitations are each described like states of open strings with identical values of \alpha' M^2. The value of \alpha' M^2 for the closed string state is four times that value.

How come?


Equation (14.78):

\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2

Mass M is two times that value (M = 2 M_L or M = 2 M_R). So M^2 is four times.

— Me@2015.07.14 12:47 PM


Not necessarily so.

— Me@2015.07.16 08:14 AM


Instead, it is just due to this definition.

— Me@2015.07.30 09:19 AM


p.322 “In the closed string theory the value of the mass-squared is given by…”

Equation (14.78) is actually a definition of the mass-squared of a closed string state.

Also, consider

p.322 “As befits closed strings there is also the level-matching condition \alpha_0^- = \bar \alpha_0^- on the states. This condition guarantees that the left and right sectors give identical contributions to the mass-squared: \alpha' M_L^2 = \alpha' M_R^2.”


\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2

\alpha' M^2 = 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2

— Me@2018.03.02 11:05 AM



2018.03.02 Friday (c) All rights reserved by ACHK