Problem 14.4b1.2

Closed string degeneracies | A First Course in String Theory

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(b) State the values of \alpha' M^2 and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

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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}: ~~~~\{ \alpha_{-1}^I b_{\frac{-1}{2}}^J, b_{\frac{-3}{2}}^I, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle,
\alpha'M^2=2, ~~N^\perp = \frac{5}{2}: ~~~~\{\alpha_{-2}^I b_{\frac{-1}{2}}^J, \alpha_{-1}^I \alpha_{-1}^J b_{\frac{-1}{2}}^K, \alpha_{-1}^I b_{\frac{-3}{2}}^J, \alpha_{-1}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M, ...\}
\{ ..., b_{\frac{-5}{2}}^I, b_{\frac{-3}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M b_{\frac{-1}{2}}^N \} |NS \rangle \otimes |p^+, \vec p_T \rangle,

For N^\perp = \frac{5}{2}, the number of states is

8^2 + \left[ \frac{(8)(7)}{2!} + 8 \right] (8) + 8^2
+ 8 + \frac{(8)(7)(6)}{3!} + 8 + (8) \left[ \frac{(8)(7)}{2!} \right] + \frac{(8)(7)(6)(5)(4)}{5!}
= 1152

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Since \alpha' M^2 = N^\perp - \frac{1}{2}, when N^\perp = \frac{5}{2}, \alpha' M^2 = 2.

Equation (14.67):

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

Equation (14.66):

f_{NS} (x) = \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8

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p.321

If we take f_{NS} (x) in (14.66) and change the sign inside each factor in the numerator

Equation (14.72):

\frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8

the _only_ effect is changing the sign of each term in the generating function whose states arise with an odd number of fermions.

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\prod_{n=1}^\infty \left( \frac{1}{1-x^n} \right)^8 is the boson contribution.

\prod_{n=1}^\infty \left( 1+x^{n-\frac{1}{2}} \right)^8 is the fermion contribution.

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Turning Equation (14.67) into

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

is equivalent to turning all \sqrt{x} into - \sqrt{x}:

f_{NS?} (x)
= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8

Me@2015.08.29 12:49 PM: Somehow, \sqrt{x} represents “contribution from fermions”.

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Me@2015.08.29 12:50 PM: If you still cannot understand, try replace all \sqrt{x} with y.

f_{NS+} (x) = \frac{1}{2} \left( f_{NS} - f_{NS?} \right)

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

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

f_{NS+} (x)
= \frac{1}{2} \left( f_{NS} - f_{NS?} \right)
= \frac{1}{\sqrt{x}} + 36 \sqrt{x} + 402 x \sqrt{x} + ...

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It is _not_ correct. Just consider it as \left(\sqrt{x} \to -\sqrt{x} \right) is not correct, since the beginning factor \frac{1}{\sqrt{x}} is not considered yet.

Instead, we should present in the following way:

f_{NS} (x)
= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= \frac{1}{\sqrt{x}} g_{NS}(x)
= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...

g (\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 1 + 8 \, \sqrt{x} + 36 \, x + 128 \, x^{\frac{3}{2}} + 402 \, x^{2} + 1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...

g (-\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 1 -8 \, \sqrt{x} + 36 \, x -128 \, x^{\frac{3}{2}} + 402 \, x^{2} -1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...

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g (\sqrt{x}) - g (-\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 - \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 16 \, \sqrt{x} + 256 \, x^{\frac{3}{2}} + 2304 \, x^{\frac{5}{2}} + 15360 \, x^{\frac{7}{2}} + ...

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f_{NS+}(x) = \frac{1}{2 \sqrt{x}} \left[ g (\sqrt{x}) - g (-\sqrt{x}) \right]
= \frac{1}{2 \sqrt{x}} \left[ \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 - \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8 \right]
= \frac{1}{2 \sqrt{x}} \left[ 16 \, \sqrt{x} + 256 \, x^{\frac{3}{2}} + 2304 \, x^{\frac{5}{2}} + 15360 \, x^{\frac{7}{2}} + ... \right]
= 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ...

— Me@2018-05-08 08:50:32 PM

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2018.05.08 Tuesday (c) All rights reserved by ACHK