# Problem 14.3b5

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 (14.37),

$M^2 = \frac{1}{\alpha'} \left( - \frac{1}{2} + N^\perp \right)$

$N^\perp = \frac{5}{2}$

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$\{ b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^L b_{-1/2}^M,$

$b_{-3/2}^I b_{-1/2}^L b_{-1/2}^M,$

$b_{-5/2}^I,$

$\alpha_{-1}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^M,$

$\alpha_{-1}^I b_{-3/2}^J,$

$\alpha_{-2}^I b_{-1/2}^J,$

$\alpha_{-1}^I \alpha_{-1}^J b_{-1/2}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

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There are $\frac{8 \times 7 \times 6 \times 5 \times 4}{5!} = {8 \choose 5} = 56$ states for

$\{ b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^L b_{-1/2}^M |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times \frac{8 \times 7}{2} = 224$ states for

$\{ b_{-3/2}^I b_{-1/2}^L b_{-1/2}^M \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8$ states for

$\{ b_{-5/2}^I \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times {8 \choose 3} = 448$ states for

$\{ \alpha_{-1}^I b_{-1/2}^J b_{-1/2}^K b_{-1/2}^M \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times 8 = 64$ states for

$\{ \alpha_{-1}^I b_{-3/2}^J \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times 8 = 64$ states for

$\{ \alpha_{-2}^I b_{-1/2}^J \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $\left( \frac{8 \times 7}{2!} + 8 \right) \times 8 = 288$ states for

$\{ \alpha_{-1}^I \alpha_{-1}^J b_{-1/2}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

So the total number of states is 56 + 224 + 8 + 448 + 64 + 64 + 288 = 1152.

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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@2018-02-16 03:22:13 PM

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