Acta Physica Sinica, Volume. 69, Issue 10, 101101-1(2020)
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Fig. 3. The equivalence of a closed string tree-level cylinder amplitude and an open string one-loop annulus amplitude
Fig. 4. A virtual open string circulating a closed time loop can be viewed as a pair of virtual open stringand virtual anti open string creating and annihilating.
Table 1. The equations needed to determine the corresponding eigenvalues for
.View tableView in ArticleTable 1. The equations needed to determine the corresponding eigenvalues for
.p 本征值满足的关系 0 $\lambda = 1$ 1 $\lambda_{0} + \lambda^{-1}_{0} = {\rm{tr} } { {w}}$ 2 $\lambda_{0} + \lambda^{-1}_{0} = {\rm{tr} } { {w}} - 1$ $\lambda = 1$ 3 $\displaystyle\sum_{\alpha =0}^{1} (\lambda_{\alpha} + \lambda^{-1}_{\alpha}) = {\rm{tr} } { {w}}, \;\;\displaystyle\sum_{\alpha =0}^{1} (\lambda^{2}_{\alpha} + \lambda^{-2}_{\alpha}) = {\rm{tr} } { {w}}^{2}$ 4 $\displaystyle\sum_{\alpha =0}^{1} (\lambda_{\alpha} + \lambda^{-1}_{\alpha}) = {\rm{tr} } { {w}} - 1, \;\;\displaystyle\sum_{\alpha =0}^{1} (\lambda^{2}_{\alpha} + \lambda^{-2}_{\alpha}) = {\rm{tr} } { {w}}^{2} - 1, \, \lambda = 1$ 5 $\displaystyle\sum_{\alpha =0}^{2} (\lambda_{\alpha} + \lambda^{-1}_{\alpha}) = {\rm{tr} } { {w}}, \;\;\displaystyle\sum_{\alpha =0}^{2} (\lambda^{2}_{\alpha} + \lambda^{-2}_{\alpha}) = {\rm{tr} } { {w}}^{2}, \;\;\displaystyle\sum_{\alpha =0}^{2} (\lambda^{3}_{\alpha} + \lambda^{-3}_{\alpha}) = {\rm{tr} } { {w}}^{3}$ 6 $\displaystyle\sum_{\alpha =0}^{2} (\lambda_{\alpha} + \lambda^{-1}_{\alpha}) = {\rm{tr} } { {w}} - 1, \;\;\displaystyle\sum_{\alpha =0}^{2} (\lambda^{2}_{\alpha} + \lambda^{-2}_{\alpha}) = {\rm{tr} } { {w}}^{2} - 1, $ $\displaystyle\sum_{\alpha =0}^{2} (\lambda^{3}_{\alpha} + \lambda^{-3}_{\alpha}) = {\rm{tr} } { {w}}^{3} - 1,$ $ \lambda = 1$
Table 2. The
-terms in the square bracket in Eq. (44 ) and their simplification.View tableView in ArticleTable 2. The
-terms in the square bracket in Eq. (44 ) and their simplification.p 振幅表达式(( 44 )式)方括号中$\theta$ 0 $\theta^{4}_{3} (0 |it) - \theta^{4}_{4} (0 |it) - \theta^{4}_{2} (0 | it) = 2\, \theta^{4}_{1} (0 | it) = 0$ 1 或 2 $\theta^{3}_{3} (0 |it) \theta_{3} (\nu_{0} |it) - \theta^{3}_{4} (0 |it) \theta_{4} (\nu_{0} |it) - \theta^{3}_{2} (0 |it) \theta_{2} (\nu_{0} | it) = 2\, \theta^{4}_{1} \left(\left.\dfrac{\nu_{0}}{2}\right| it \right)$ 3 或 4 $\theta^{2}_{3} (0 |it) \theta_{3} (\nu_{0} |it) \theta_{3} (\nu_{1} |it)- \theta^{2}_{4} (0 |it) \theta_{4} (\nu_{0} |it) \theta_{4} (\nu_{1} |it)- \theta^{2}_{2} (0 |it) \theta_{2} (\nu_{0} | it)\theta_{2} (\nu_{1} | it)$ $= 2\, \theta^{2}_{1} \left(\left.\dfrac{\nu_{0} + \nu_{1} }{2}\right|it \right) \theta^{2}_{1} \left(\left.\dfrac{\nu_{0} - \nu_{1} }{2}\right|it \right)$ 5 或 6 $\theta_{3} (0 |it) \theta_{3} (\nu_{0} |it) \theta_{3} (\nu_{1} |it) \theta_{3} (\nu_{2} |it) - \theta_{4} (0 |it) \theta_{4} (\nu_{0} |it) \theta_{4} (\nu_{1} |it) \theta_{4} (\nu_{2} |it)$ $ - \theta_{2} (0 |it) \theta_{2} (\nu_{0} | it)\theta_{2} (\nu_{1} | it) \theta_{2} (\nu_{2} | it) $ $= 2\, \theta_{1} \left(\left.\dfrac{\nu_{0} + \nu_{1} + \nu_{2} }{2}\right| it \right) \theta_{1} \left(\left.\dfrac{\nu_{0} - \nu_{1} + \nu_{2} }{2}\right| it \right) \theta_{1} \left(\left.\dfrac{\nu_{0} + \nu_{1} - \nu_{2} }{2}\right| it \right)\theta_{1} \left(\left.\dfrac{\nu_{0} - \nu_{1} - \nu_{2} }{2}\right| it \right) $
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Jian-Xin Lu, Nan Zhang.
Paper Information
Received: Jan. 6, 2020
Accepted: --
Published Online: Nov. 30, 2020
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