# 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?

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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

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Not necessarily so.

— Me@2015.07.16 08:14 AM

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Instead, it is just due to this definition.

— Me@2015.07.30 09:19 AM

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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: $\alpha' M_L^2 = \alpha' M_R^2$.”

Then $\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2$ $\alpha' M^2 = 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2$

— Me@2018.03.02 11:05 AM

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