Problem 14.4b1.2

Closed string degeneracies | A First Course in String Theory

(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?

~~~

…

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

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

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):

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

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.

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

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

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”.

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

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

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

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

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

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:

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

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

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

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

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

— Me@2018-05-08 08:50:32 PM

Physics Town

continues

Problem 14.4b1.2

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