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The Heterotic string state is a tensoring of the bosonic string left-moving state and the Type II string right-moving state. Therefore, I expect the spectrum to be: $$\begin{array}{*{20}{c}} \hline & {{\rm{Sector}}}&{{\rm{Spectrum}}}&{{\rm{Massless fields}}}& \\ \hline & {{\rm{Bosonic}} - {\rm{R}}}&{{\bf{1}}{{\bf{6}}_v} \otimes {{\bf{8}}_s} = {{\bf{8}}_v} \otimes {{\bf{8}}_v} \otimes {{\bf{8}}_s}}&?\\ \hline & {{\rm{Bosonic}} - {\rm{NS}}}&{{{\bf{8}}_v} \otimes {{\bf{8}}_v} \otimes {{\bf{8}}_v}}&?& \hline \end{array}$$

  1. However, how does one calculate the massless fields of the Type I string theory using the spectrum of the type I string theory?

  2. Furthermore, to calculate the mass spectrum of the Heterotic string, does one simply add the number operator of the bosonic string to that of the type II string and the same for the normal ordering constant? i.e. is it true that

$$\begin{array}{l} m = \sqrt {\frac{{2\pi T}}{{{c_0}}}\left( {B + {{\tilde N}_{II}} - {a_B} - {{\tilde a}_{II}}} \right)} \\ {\rm{ }} = \sqrt {\frac{{2\pi T}}{{{c_0}}}\left( {B + {{\tilde N}_{II}} - 1 - {{\tilde a}_{II}}} \right)} \end{array}$$

Edit: I found the answer to Question 2. Check the answers section. I have answered my own question.

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Maybe you find this recent TRF article about heterotic string theory interesting too. –  Dilaton May 18 '13 at 17:10

1 Answer 1

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I just found that my question (2) is trivially easy and that indeed it is:
    m = \sqrt {\frac{{2\pi T}}{{{c_0}}}\left( {B + {{\tilde N}_{II}} - {a_B} - {{\tilde a}_{II}}} \right)} \\
    {\rm{  }} = \sqrt {\frac{{2\pi T}}{{{c_0}}}\left( {B + {{\tilde N}_{II}} - 1 - {{\tilde a}_{II}}} \right)} 

This is because:

$$m=\sqrt{\frac{2\pi T}{c_0}\left(N+\tilde N-a-\tilde a\right)}$$

*Edit: * Ooops. I was wrong, The correct answer is (in the RR sector): $$m = \frac{2\pi T\ell_s}{c_0^2}\sqrt {{N + \tilde N+2}} $$ In the N-SN-S sector, the correct answer is: $$m = \frac{2\pi T\ell_s}{c_0^2}\sqrt {{N +\tilde N - 2}} $$ In the RN-S/N-SR sectors, $$m = \frac{2\pi T\ell_s}{c_0^2}\sqrt {{N+\tilde N}} $$

This is because its Ramond sector normal-ordering constant is $-1,$ while its Neveu-Schwarz sector normal-ordering constant is $1.$

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