I'm working through Polchinski's little book, in which he describes how to get string amplitudes in a somewhat heuristic manner, without the machinery introduced in the big book. These lectures are wonderful, but problem is that he just states a lot of results without the details. For instance, I'm interested in the three-gauge bosons amplitude (equation 10.12 and equation 6.5.15 in the big book), but I'm getting crazy with an important relative missing -1/2 factor in the cubic term. I attach the calculation, if anybody can help me I would be extremely grateful.

The amplitude is

\begin{align} S(k_1,k_2,k_3)&=-\frac{g_o}{\sqrt{8\alpha'^5}}|y_{12}y_{13}y_{23}|\Bigg<\prod_{1,2,3}: \:e_i\cdot\left(\partial_{y_i}X(y_i)\right)e^{ik_i\cdot X(y_i)} :\:\Bigg>. \end{align} For short let's define the notation

\begin{align} &[e_1e^2]\equiv (e_1\cdot \dot{X}(y_1)) \: e^{ik_2\cdot X(y_2)}=[e^2e_1]=-2i\alpha'\partial_1\ln|y_{12}|e_1\cdot k_2=2i\alpha'\frac{e_1\cdot k_2}{|y_{12}|}\:\:\:(\text{with a $-1$ if } y_1>y_2)\\ &[e_1e_2]\equiv (e_1\cdot \dot{X}(y_1)) \:(e_2\cdot \dot{X}(y_2)) =[e_2e_1]=-2\alpha'e_1\cdot e_2\partial_1\partial_2\ln |y_{12}|=-2\alpha'\frac{e_1\cdot e_2}{ |y_{12}|^2}\\ &[e^1e^2]\equiv e^{ik_1\cdot X(y_1)} \: e^{ik_2\cdot X(y_2)}=[e^2e^1]=|y_{12}|^{2\alpha'k_1\cdot k_2}, \end{align}

where the products are understood to be contracted, and up to momentum-conserving Delta functions. The $[e^i e^j]$ will all be one because of the massless condition. Thus \begin{align} S(k_1,k_2,k_3)=-\frac{g_o}{\sqrt{8\alpha'^5}}|y_{12}y_{13}y_{23}|&\Bigg([e_1e^2][e_2e^3][e_3e^1]+[e_1e^3][e_3e^2][e_2e^1]\nonumber\\ &+[e_1e_2]\left(\:[e^1e^3][e_3e^2]+[e^1e_3][e^3e^2]\:\right)+(1,3)+(2,3)\Bigg), \end{align}

with the repetitions $(1,3)$ and $(2,3)$ applying only to the second term. So we have,

\begin{align} S(k_1,k_2,k_3)=-\frac{g_o}{\sqrt{8\alpha'^5}}|y_{12}y_{13}y_{23}|&\Bigg((2i\alpha')^3\frac{e_1\cdot k_2}{|y_{12}|}\frac{e_2\cdot k_3}{|y_{23}|}\frac{-e_3\cdot k_1}{|y_{13}|}+(2i\alpha')^3\frac{e_1\cdot k_3}{|y_{13}|}\frac{-e_3\cdot k_2}{|y_{23}|}\frac{-e_2\cdot k_1}{|y_{21}|}\nonumber\\ &-2\alpha'\frac{e_1\cdot e_2}{ |y_{12}|^2}\left(\:2i\alpha'\frac{-e_3\cdot k_2}{|y_{23}|}+2i\alpha'\frac{-e_3\cdot k_1}{|y_{13}|}\:\right)+(1,3)+(2,3)\Bigg)\\ =-\frac{g_o}{\sqrt{8\alpha'^5}}|y_{12}y_{13}y_{23}|&\Bigg(\frac{8i\alpha'^3}{|y_{12}y_{13}y_{23}|}\left(e_1\cdot k_2\:e_2\cdot k_3\:e_3\cdot k_1-e_1\cdot k_3\:e_3\cdot k_2\:e_2\cdot k_1\right)\nonumber\\ &+4i\alpha'^2\frac{e_1\cdot e_2}{ |y_{12}|^2}\:e_3\cdot\left(\:\frac{ k_2}{|y_{23}|}+\frac{ k_1}{|y_{13}|}\:\right)+(1,3)+(2,3)\Bigg). \end{align}

The dot products in the first factor sum up to

\begin{equation} e_1\cdot k_2\:e_2\cdot k_3\:e_3\cdot k_1-e_1\cdot k_3\:e_3\cdot k_2\:e_2\cdot k_1= \frac{1}{4}e_1\cdot(k_2-k_3)\:e_2\cdot(k_3-k_1)\:e_3\cdot(k_1-k_2), \end{equation}

and also, for $y_1=0, y_2=1, y_3=\infty$,

\begin{align} e_3\cdot\left(|y_{13}|k_2+|y_{23}|k_1 \right)= e_3\cdot\left(y_3(k_2+k_1)-y_1k_2-y_2k_1\right)=-e_3\cdot k_1=-\frac{1}{2}e_3\cdot(k_1-k_2), \end{align}


\begin{align} S(k_1,k_2,k_3) =-\frac{ig_o\alpha'^2}{\sqrt{8\alpha'^5}}&\Bigg(2\alpha'e_1\cdot(k_2-k_3)\:e_2\cdot(k_3-k_1)\:e_3\cdot(k_1-k_2)\nonumber\\ &-2(e_1\cdot e_2)\:e_3\cdot(k_1-k_2)+(1,3)+(2,3)\Bigg)\\ =\frac{2ig_o\alpha'^2}{\sqrt{8\alpha'^5}}&\Bigg(-\alpha'e_1\cdot(k_2-k_3)\:e_2\cdot(k_3-k_1)\:e_3\cdot(k_1-k_2)\nonumber\\ &+(e_1\cdot e_2)\:e_3\cdot(k_1-k_2)+(1,3)+(2,3)\Bigg)\\ =\frac{ig_o}{\sqrt{2\alpha'}}&\Bigg(-\alpha'e_1\cdot(k_2-k_3)\:e_2\cdot(k_3-k_1)\:e_3\cdot(k_1-k_2)\nonumber\\ &+(e_1\cdot e_2)\:e_3\cdot(k_1-k_2)+(1,3)+(2,3)\Bigg) \end{align}

Apparently, I missed a factor of -1/2 in the first term, the prefactor should be $\alpha'/2$. Also Green-Schwarz-Witten report the same amplitude as Polchinski, but nowhere in the literature I have seen the detailed calculation. The prefactor is relevant for the efective gauge action.


1 Answer 1


I have not gone through your calculation, but this is my detailed calculation.

We first consider the matter part: \begin{align} S_3^{X \mu\nu\sigma} \left\langle : \dot{X}^\mu e^{ik_1\cdot X}(y_1) : \, : \dot{X}^\nu e^{ik_2\cdot X}(y_2) : \, : \dot{X}^\sigma e^{ik_3\cdot X}(y_3) : \, \right \rangle \end{align} From (6.2.36) we get \begin{align} S_3^{X\mu\nu\sigma} =& i C_{D_2}^X (2\pi)^{26} \delta^{26}(\sum_i k_i) |y_{12}|^{2\alpha' k_1\cdot k_2}|y_{13}|^{2\alpha' k_1\cdot k_3}|y_{23}|^{2\alpha' k_2\cdot k_3} \Big[ v^\mu (y_1) v^\nu (y_2) v^\sigma (y_3) \nonumber\\ &+ v^\mu (y_1) \langle q^\nu(y_2) q^\sigma (y_3) \rangle + v^\nu (y_2) \langle q^\mu(y_1) q^\sigma (y_3) \rangle + v^\sigma (y_3) \langle q^\mu(y_1) q^\nu (y_2) \rangle \Big] \nonumber\\ & \times \mathrm{tr} {\lambda^{a_1} \lambda^{a_2} \lambda^{a_3}} + (k_2,a_2,\epsilon_2) \leftrightarrow (k_3,a_3,\epsilon_3) \nonumber\\ =& i C_{D_2}^X (2\pi)^{26} \delta^{26}(\sum_i k_i) |y_{12}|^{2\alpha' k_1\cdot k_2}|y_{13}|^{2\alpha' k_1\cdot k_3}|y_{23}|^{2\alpha' k_2\cdot k_3} \nonumber\\ &\times \Bigg\{(-2i\alpha')^3 \left( \frac{k_2^\mu}{y_{12}} + \frac{k_3^\mu}{y_{13}} \right) \left( \frac{k_1^\nu}{y_{21}} + \frac{k_3^\nu}{y_{23}} \right)\left( \frac{k_1^\sigma}{y_{31}} + \frac{k_2^\sigma}{y_{32}} \right) \nonumber\\ & + (-2\alpha')(-2i\alpha') \left[ \left( \frac{k_2^\mu}{y_{12}} + \frac{k_3^\mu}{y_{13}} \right) \frac{\eta^{\nu\sigma}}{y_{23}^2} + \left( \frac{k_1^\nu}{y_{21}} + \frac{k_3^\nu}{y_{23}} \right) \frac{\eta^{\mu\nu}}{y_{13}^2} +\left( \frac{k_1^\sigma}{y_{31}} + \frac{k_2^\sigma}{y_{32}} \right) \frac{\eta^{\mu\sigma}}{y_{12}^2} \right]\Bigg\}\nonumber\\ & \times \mathrm{tr} {\lambda^{a_1} \lambda^{a_2} \lambda^{a_3}} + (k_2,a_2,\epsilon_2) \leftrightarrow (k_3,a_3,\epsilon_3) \end{align} The first thing we note is that because of momentum conservation and mass-shell condition \begin{align} 0= k_1^2 =(-k_2-k_3)^2 = k_2^2+k_3^2 + 2 k_1\cdot k_2 \Rightarrow k_1\cdot k_2 =0 \end{align} We now add the ghost contribution $\langle ccc\rangle = C_{D_2}^g y_{12} y_{13} y_{23}$, the polarisation vectors, the cosmological term $e^{-\lambda}$ and the normalisation of the boson vertex operators, $-ig_0/\sqrt{2\alpha'}$ see (3.6.26), and we find \begin{align} S_{D_2}^{ggg} =& S_{D_2} (k_1,a_1, \epsilon_1; k_2,a_2, \epsilon_2; k_3,a_3, \epsilon_3) \nonumber\\ =& \left(-\frac{ig_0}{\sqrt{2\alpha'}}\right)^3 4i\alpha'^2 C_{D_2}^X C_{D_2}^g e^{-\lambda} \epsilon_{1\mu} \epsilon_{2\nu} \epsilon_{3\sigma} (2\pi)^{26} \delta^{26}(\sum_i k_i) y_{12} y_{13} y_{23}\nonumber\\ &\times \Bigg[ 2\alpha' \left( \frac{k_2^\mu}{y_{12}} + \frac{k_3^\mu}{y_{13}} \right) \left( \frac{k_1^\nu}{y_{21}} + \frac{k_3^\nu}{y_{23}} \right)\left( \frac{k_1^\sigma}{y_{31}} + \frac{k_2^\sigma}{y_{32}} \right) \nonumber\\ &+ \left( \frac{k_2^\mu}{y_{12}} + \frac{k_3^\mu}{y_{13}} \right) \frac{\eta^{\nu\sigma}}{y_{23}^2} + \left( \frac{k_1^\nu}{y_{21}} + \frac{k_3^\nu}{y_{23}} \right) \frac{\eta^{\mu\nu}}{y_{13}^2} +\left( \frac{k_1^\sigma}{y_{31}} + \frac{k_2^\sigma}{y_{32}} \right) \frac{\eta^{\mu\sigma}}{y_{12}^2} \Bigg]\nonumber\\ & \times \mathrm{tr} {\lambda^{a_1} \lambda^{a_2} \lambda^{a_3}} + (k_2,a_2,\epsilon_2) \leftrightarrow (k_3,a_3,\epsilon_3) \end{align} Let us first simplify the pre-factor. Use (6.4.14), i.e. $C_{D_2}^X C_{D_2}^g e^{-\lambda} = C_{D_2} = 1/\alpha' g_0^2$ to write \begin{align} \left(-\frac{ig_0}{\sqrt{2\alpha'}}\right)^3 4i\alpha'^2 C_{D_2}^X C_{D_2}^g e^{-\lambda} = -\frac{2 \alpha' g_0^3}{\sqrt{2\alpha'}} C_{D_2}^X C_{D_2}^g e^{-\lambda} = -\frac{2g_0}{\sqrt{2\alpha'}} =-2 g_0' \end{align} in the last line we have used (6.5.14), i.e. $g_0'= g_0/\sqrt{2\alpha'}$. The three boson amplitude thus becomes \begin{align} S_{D_2}^{ggg} =& S_{D_2} (k_1,a_1, \epsilon_1; k_2,a_2, \epsilon_2; k_3,a_3, \epsilon_3) \nonumber\\ =& -2g_0' \epsilon_{1\mu} \epsilon_{2\nu} \epsilon_{3\sigma} (2\pi)^{26} \delta^{26}(\sum_i k_i) y_{12} y_{13} y_{23} \Bigg[ 2\alpha' \left( \frac{k_2^\mu}{y_{12}} + \frac{k_3^\mu}{y_{13}} \right) \left( \frac{k_1^\nu}{y_{21}} + \frac{k_3^\nu}{y_{23}} \right)\left( \frac{k_1^\sigma}{y_{31}} + \frac{k_2^\sigma}{y_{32}} \right) \nonumber\\ &+ \left( \frac{k_2^\mu}{y_{12}} + \frac{k_3^\mu}{y_{13}} \right) \frac{\eta^{\nu\sigma}}{y_{23}^2} + \left( \frac{k_1^\nu}{y_{21}} + \frac{k_3^\nu}{y_{23}} \right) \frac{\eta^{\mu\nu}}{y_{13}^2} +\left( \frac{k_1^\sigma}{y_{31}} + \frac{k_2^\sigma}{y_{32}} \right) \frac{\eta^{\mu\sigma}}{y_{12}^2} \Bigg]\nonumber\\ & \times \mathrm{tr} {\lambda^{a_1} \lambda^{a_2} \lambda^{a_3}} + (k_2,a_2,\epsilon_2) \leftrightarrow (k_3,a_3,\epsilon_3) \end{align} In the remainder of the calculation we will repeatedly use the fact that $k_i\cdot \epsilon_i=0$ for $i=1,2,3$. This actually means that we can simply ignore any terms that have a $k_1^\mu, k_2^\nu$ or a $k_3^\sigma$ as these will be contracted with the polarisation vectors $\epsilon_{1\mu},\epsilon_{2\nu}$ and $\epsilon_{3\sigma}$ respectively. Consider first the terms linear in momentum and take the first such terms \begin{align} & y_{12} y_{13} y_{23} \left( \frac{k_2^\mu}{y_{12}} + \frac{k_3^\mu}{y_{13}} \right) \frac{\eta^{\nu\sigma}}{y_{23}^2} = \frac{\eta^{\nu\sigma}}{y_{23}}(k_2^\mu y_{13} +k_3^\mu y_{12}) \nonumber\\ =& \frac{\eta^{\nu\sigma}}{2y_{23}}\big[k_2^\mu y_{13} +(-k_1^\mu-k_3^\mu ) y_{13}+k_3^\mu y_{12}+(-k_1^\mu-k_2^\mu) y_{12}\big] \nonumber\\ =& \frac{\eta^{\nu\sigma}}{2y_{23}} (k_{23}^\mu y_{13} +k_{32}^\mu y_{12} ) = \frac{k_{23}^\mu \eta^{\nu\sigma}}{2y_{23}} (y_{13}-y_{12})= \frac{k_{23}^\mu \eta^{\nu\sigma}}{2y_{23}} y_{23} = \frac{1}{2} k_{23}^\mu \eta^{\nu\sigma} \end{align} Contracting with the polarisation vectors $\epsilon_1^\mu \epsilon_2^\nu \epsilon_3^\sigma$ this gives a contribution $\frac{1}{2} (\epsilon_1\cdot k_{23} )( \epsilon_2\cdot \epsilon_3)$. The two other terms linear in the momenta give similar contributions so that we can write \begin{align} S_{D_2}^{ggg [1]} =& -g_0' (2\pi)^{26} \delta^{26}(\sum_i k_i) \times \big[ (\epsilon_1\cdot k_{23} )(\epsilon_2\cdot \epsilon_3) + (\epsilon_2\cdot k_{31} )(\epsilon_3\cdot \epsilon_1)+ (\epsilon_3\cdot k_{12} )(\epsilon_1\cdot \epsilon_2) \big]\nonumber\\ &\qquad \times \mathrm{tr} {\lambda^{a_1} \lambda^{a_2} \lambda^{a_3}} + (k_2,a_2,\epsilon_2) \leftrightarrow (k_3,a_3,\epsilon_3) \nonumber\\ &= -g_0' (2\pi)^{26} \delta^{26}(\sum_i k_i) \Bigg\{ \big[ (\epsilon_1\cdot k_{23} )(\epsilon_2\cdot \epsilon_3) + (\epsilon_2\cdot k_{31} )(\epsilon_3\cdot \epsilon_1) \nonumber\\ & + (\epsilon_3\cdot k_{12} )(\epsilon_1\cdot \epsilon_2) \big] \mathrm{tr} {\lambda^{a_1} \lambda^{a_2} \lambda^{a_3}} \nonumber\\ &+ \big[ (\epsilon_1\cdot k_{32} )(\epsilon_3\cdot \epsilon_2) + (\epsilon_3\cdot k_{21} )(\epsilon_2\cdot \epsilon_1)+ (\epsilon_2\cdot k_{13} )(\epsilon_1\cdot \epsilon_3) \big] \mathrm{tr} {\lambda^{a_1} \lambda^{a_3} \lambda^{a_2}} \Bigg\} \end{align} and finally for the linear term \begin{align} S_{D_2}^{ggg [1]}&= -g_0' (2\pi)^{26} \delta^{26}(\sum_i k_i) \nonumber\\ &\times \big[ (\epsilon_1\cdot k_{23} )(\epsilon_2\cdot \epsilon_3) + (\epsilon_2\cdot k_{31} )(\epsilon_3\cdot \epsilon_1)+ (\epsilon_3\cdot k_{12} )(\epsilon_1\cdot \epsilon_2) \big] \mathrm{tr} {\lambda^{a_1} [ \lambda^{a_2}, \lambda^{a_3}}] \end{align} Let us now focus on the term cubic in the momenta. We have \begin{align} &\left( \frac{k_2^\mu}{y_{12}} + \frac{k_3^\mu}{y_{13}} \right) \left( \frac{k_1^\nu}{y_{21}} + \frac{k_3^\nu}{y_{23}} \right)\left( \frac{k_1^\sigma}{y_{31}} + \frac{k_2^\sigma}{y_{32}} \right) \nonumber\\ =& \frac{k_2^\mu y_{13} +k_3^\mu y_{12}}{y_{12} y_{13}} \frac{k_1^\nu y_{23} +k_3^\nu y_{21}}{y_{21} y_{23}} \frac{k_1^\sigma y_{32} +k_2^\sigma y_{31}}{y_{31} y_{32}} \nonumber\\ =& \frac{[k_2^\mu y_{13} +(-k_1^\mu-k_2^\mu) y_{12}] [ k_1^\nu y_{23} +(-k_1^\nu-k_2^\nu) y_{21}][ k_1^\sigma y_{32} +(-k_1^\sigma -k_3^\sigma) y_{31} ]}{y_{12} y_{13} y_{21}y_{23} y_{31} y_{32}}\nonumber\\ =&- \frac{k_2^\mu(y_{13}-y_{12}) k_1^\nu(y_{23} -y_{21}) ( k_1^\sigma (y_{32} - y_{31})} {y_{12}^2 y_{13}^2 y_{23}^2}\nonumber\\ =& \frac{k_2^\mu k_1^\nu k_1^\sigma y_{23} y_{13} y_{12}}{ y_{12}^2 y_{13}^2 y_{23}^2} = \frac{k_2^\mu k_1^\nu k_1^\sigma }{ y_{12} y_{13} y_{23}} \end{align} We have repeatedly used momentum conservation and the fact that $\epsilon_1\cdot k_1 = \epsilon_2\cdot k_2 =\epsilon_3\cdot k_3 =0$. We now rewrite \begin{align} &k_2^\mu k_1^\nu k_1^\sigma = \left(\frac{1}{2}\right)^3 [k_2^\mu + (-k_1^\mu-k_3^\mu) ] [k_1^\nu + (-k_2^\nu-k_3^\nu) ] [k_1^\sigma + (-k_2^\sigma-k_3^\sigma) ]\nonumber\\ & = \frac{1}{8} k_{23}^\mu k_{13}^\nu k_{12}^\sigma \end{align} Bringing it all together, we find for the cubic terms \begin{align} S_{D_2}^{ggg [3]}= & -2g_0' (2\pi)^{26} \delta^{26}(\sum_i k_i) y_{12} y_{13} y_{23} \epsilon_1^\mu \epsilon_2^\nu \epsilon_3^\sigma 2 \alpha' \frac{k_{23}^\mu k_{13}^\nu k_{12}^\sigma}{8y_{12} y_{13} y_{23} } \mathrm{tr} {\lambda^{a_1} \lambda^{a_2} \lambda^{a_3}} \nonumber\\ & + (k_2,a_2,\epsilon_2) \leftrightarrow (k_3,a_3,\epsilon_3) \nonumber\\ =& -g_0' (2\pi)^{26} \delta^{26}(\sum_i k_i)\frac{\alpha'}{2} (\epsilon_1\cdot k_{23})(\epsilon_2\cdot k_{13})(\epsilon_3\cdot k_{12}) \mathrm{tr} {\lambda^{a_1} \lambda^{a_2} \lambda^{a_3}} \nonumber\\ & + (k_2,a_2,\epsilon_2) \leftrightarrow (k_3,a_3,\epsilon_3) \nonumber\\ =& -g_0' (2\pi)^{26} \delta^{26}(\sum_i k_i)\frac{\alpha'}{2} \Big[ (\epsilon_1\cdot k_{23})(\epsilon_2\cdot k_{13})(\epsilon_3\cdot k_{12}) \mathrm{tr} {\lambda^{a_1} \lambda^{a_2} \lambda^{a_3}} \nonumber\\ & + (\epsilon_1\cdot k_{32})(\epsilon_3\cdot k_{12})(\epsilon_2\cdot k_{13}) \mathrm{tr} {\lambda^{a_1} \lambda^{a_3} \lambda^{a_2}} \Big] \nonumber\\ =& -g_0' (2\pi)^{26} \delta^{26}(\sum_i k_i)\frac{\alpha'}{2} (\epsilon_1\cdot k_{23})(\epsilon_2\cdot k_{13})(\epsilon_3\cdot k_{12}) \mathrm{tr} {\lambda^{a_1}[ \lambda^{a_2}, \lambda^{a_3}} ] \end{align} Bringing the cubic and the linear terms together we thus find \begin{align} S_{D_2} (k_1,a_1, \epsilon_1&; k_2,a_2, \epsilon_2; k_3,a_3, \epsilon_3) = -g_0' (2\pi)^{26} \delta^{26}(\sum_i k_i) \nonumber\\ & \times \Big[ (\epsilon_1\cdot k_{23} )(\epsilon_2\cdot \epsilon_3) + (\epsilon_2\cdot k_{31} )(\epsilon_3\cdot \epsilon_1)+ (\epsilon_3\cdot k_{12} )(\epsilon_1\cdot \epsilon_2) \nonumber\\ & \quad + \frac{\alpha'}{2} (\epsilon_1\cdot k_{23})(\epsilon_2\cdot k_{13})(\epsilon_3\cdot k_{12})\Big] \mathrm{tr} {\lambda^{a_1}[ \lambda^{a_2}, \lambda^{a_3}} ] \end{align} which is (6.5.15), up to a factor $-i$, but this might just be a normalisation issue of the boson vertex operator as $-i = i^3$.

  • $\begingroup$ I found it very useful, thank you. Though it's not in the spirit of Joe's little book (where he doesn't introduce ghosts at all) it helped me very much. $\endgroup$ Apr 25, 2020 at 7:16
  • $\begingroup$ thanks. if you go to my profile, you can find a link to more notes on Joe's (big) book. $\endgroup$ Apr 25, 2020 at 11:51

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