# Problem 14.3b6

Quick Calculation 14.4b | A First Course in String Theory

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Massive level in the open superstring

Consider the first and second excited levels of the open superstring ($\alpha' M^2 = 1$ and $\alpha' M^2 = 2$). List the states in NS sector and the states in the R sector. Confirm that you get the same number of states.

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When $\alpha' M^2 = 2$, by Equation (14.54), the possible states are

$\{ \alpha_{-2}^I, \alpha_{-1}^I \alpha_{-1}^J, d_{-1}^I d_{-1}^J \} | R_a \rangle, || ...$

$\{ \alpha_{-1}^I d_{-1}^J, d_{-2}^I \} | R_{\bar a} \rangle, || ...$

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For $\alpha_{-2}^I | R_a \rangle$, the number of states is 8.

For $\alpha_{-1}^I \alpha_{-1}^J | R_a \rangle$, the number of states is $\frac{8 \times 7}{2} + 8 = 36$.

For $d_{-1}^I d_{-1}^J | R_a \rangle$, the number of states is $\frac{8 \times 7}{2} = 28$.

For $\alpha_{-1}^I d_{-1}^J | R_{\bar a} \rangle$, the number of states is $8 \times 8 = 64$.

For $d_{-2}^I | R_{\bar a} \rangle$, the number of states is 8.

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However, since each of $a$ and ${\bar a}$ has 8 possible values, there is an additional multiple of 8.

The total number of states is $8 \left[ 8 + 36 + 28 + 64 + 8 \right]$.

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You can check this answer against Equation (14.67):

$f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

— Me@2018.02.20 10:57 AM

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