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I am stuck in deriving the vertex factor for the Feynman diagrams for the QCD Lagrangian. For the quantum Yang Mills theory we will have the following interacting Lagrangian. $$ \begin{aligned} \mathcal{L}_{\text {int }}= & -g f^{a b c}\left(\partial_\mu A_\nu^a\right) A_\mu^b A_\nu^c-\frac{1}{4} g^2\left(f^{e a b} A_\mu^a A_\nu^b\right)\left(f^{e c d} A_\mu^c A_\nu^d\right)+g f^{a b c}\left(\partial_\mu \bar{c}^a\right) A_\mu^b c^c \\ & +g A_\mu^a \bar{\psi}_i \gamma^\mu T_{i j}^a \psi_j+i g A_\mu^a T_{i j}^a\left(\phi_i^{\star} \partial_\mu \phi_j-\phi_j \partial_\mu \phi_i^{\star}\right)+g^2 \phi_i^{\star} A_\mu^a T_{i k}^a T_{k j}^b A_\mu^b \phi_j, \end{aligned}\tag{26.8} $$ For the time being consider only three gluon interaction part, which is the first term. Consider the type, polarisation and momenta of the incoming gluons are $(a,\mu,k)$, $(b,\nu,p)$,$(c,\rho,q)$. Then the vertex factor is coming to be (Schwarz, pp. 510) $$ =g f^{a b c}\left[g^{\mu \nu}(k-p)^\rho+g^{\nu \rho}(p-q)^\mu+g^{\rho \mu}(q-k)^\nu\right].\tag{26.9} $$ In the above vertex factor, I understand the appearance of $g^{\mu\nu}$ due to the polarisations but how the $(k-p)^\rho$ type of terms are coming?

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    $\begingroup$ Hint: try looking at scalar QED where these also show up. $\endgroup$ Feb 27, 2023 at 9:23

1 Answer 1

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The interaction vertex with 3-gluons is: \begin{align*} S_{I(3G)}=&- ig_s \eta_{\rho \nu} f^{a b c} \int d^4 x \int d^4 p_1 \int d^4 p_2 \int d^4 p_3\,p_{1 \mu} \widetilde{g}^{a \rho}(p_1) \widetilde{g}^{b \mu}(p_2) \widetilde{g}^{c \nu}(p_3)e^{ix_\sigma(p_1^\sigma+p_2^\sigma+p_2^\sigma)} \\ =&- i g_s \eta_{\rho \nu} f^{a b c} (2 \pi)^4 \int d^4 p_1 \int d^4 p_2 \int d^4 p_3\,\delta^{(4)}(p_1^\sigma+p_2^\sigma+p_3^\sigma) p_{1 \mu} \widetilde{g}^{a \rho}(p_1) \widetilde{g}^{b \mu}(p_2) \widetilde{g}^{c \nu}(p_3) \end{align*} A brute force chain derivative gives: \begin{align*} \frac{\delta^3 \widetilde{g}^{a \rho}(p_1) \widetilde{g}^{b \mu}(p_2) \widetilde{g}^{c \nu}(p_3)}{\delta \widetilde{g}^{a' \rho'}(p'_1) \delta \widetilde{g}^{b' \mu'}(p'_2) \delta \widetilde{g}^{c' \nu'}(p'_3)}=&\delta^a_{c'}\delta^b_{b'}\delta^c_{a'}\delta^\rho_{\nu'}\delta^\mu_{\mu'}\delta^\nu_{\rho'}\delta^{(4)}(p_1^\sigma -{p'}_3^\sigma)\delta^{(4)}(p_2^\sigma -{p'}_2^\sigma)\delta^{(4)}(p_3^\sigma -{p'}_1^\sigma) \\ &+\delta^a_{c'}\delta^b_{a'}\delta^c_{b'}\delta^\rho_{\nu'}\delta^\mu_{\rho'}\delta^\nu_{\mu'}\delta^{(4)}(p_1^\sigma -{p'}_3^\sigma)\delta^{(4)}(p_2^\sigma -{p'}_1^\sigma)\delta^{(4)}(p_3^\sigma -{p'}_2^\sigma) \\ &+\delta^a_{b'}\delta^b_{c'}\delta^c_{a'}\delta^\rho_{\mu'}\delta^\mu_{\nu'}\delta^\nu_{\rho'}\delta^{(4)}(p_1^\sigma -{p'}_2^\sigma)\delta^{(4)}(p_2^\sigma -{p'}_3^\sigma)\delta^{(4)}(p_3^\sigma -{p'}_1^\sigma) \\ &+\delta^a_{a'}\delta^b_{c'}\delta^c_{b'}\delta^\rho_{\rho'}\delta^\mu_{\nu'}\delta^\nu_{\mu'}\delta^{(4)}(p_1^\sigma -{p'}_1^\sigma)\delta^{(4)}(p_2^\sigma -{p'}_3^\sigma)\delta^{(4)}(p_3^\sigma -{p'}_2^\sigma) \\ &+\delta^a_{b'}\delta^b_{a'}\delta^c_{c'}\delta^\rho_{\mu'}\delta^\mu_{\rho'}\delta^\nu_{\nu'}\delta^{(4)}(p_1^\sigma -{p'}_2^\sigma)\delta^{(4)}(p_2^\sigma -{p'}_1^\sigma)\delta^{(4)}(p_3^\sigma -{p'}_3^\sigma) \\ &+\delta^a_{a'}\delta^b_{b'}\delta^c_{c'}\delta^\rho_{\rho'}\delta^\mu_{\mu'}\delta^\nu_{\nu'}\delta^{(4)}(p_1^\sigma -{p'}_1^\sigma)\delta^{(4)}(p_2^\sigma -{p'}_2^\sigma)\delta^{(4)}(p_3^\sigma -{p'}_3^\sigma) \end{align*} This gives terms of the form: \begin{equation*} \eta_{\rho \nu}f^{a b c} \delta^a_{c'}\delta^b_{b'}\delta^c_{a'}\delta^\rho_{\nu'}\delta^\mu_{\mu'}\delta^\nu_{\rho'}\int d^4 p_1 d^4 p_2 d^4 p_3\, p_{1 \mu}\delta^{(4)}(p_1^\sigma -{p'}_3^\sigma)\delta^{(4)}(p_2^\sigma -{p'}_2^\sigma)\delta^{(4)}(p_3^\sigma -{p'}_1^\sigma)=\eta_{\nu' \rho'} f^{c' b' a'}p'_{3 \mu'}, \end{equation*} up to some obvious permutations. Using these ones gives the total contribution of the vertex to be: \begin{align*} &\eta_{\nu' \rho'} f^{c' b' a'}p'_{3 \mu'}+\eta_{\nu' \mu'}f^{c' a' b'} p'_{3 \rho'}+\eta_{\mu' \rho'}f^{b' c' a'} p'_{2 \nu'}+\eta_{\rho' \mu'}f^{a' c' b'} p'_{1 \nu'}+\eta_{\mu' \nu'}f^{b' a' c'} p'_{2 \rho'}+\eta_{\rho' \nu'}f^{a' b' c'} p'_{1 \mu'} \\ =&g_sf^{a' b' c'}[\eta_{\mu' \nu'}(p'_3-p'_2)_{\rho'}+\eta_{\nu' \rho'}(p'_1-p'_3)_{\mu'}+\eta_{\rho' \mu'}(p'_2-p'_1)_{\nu'}] \end{align*} Which should be, ignoring the primes, the correct expression you gave (maybe up to some rearrangements...)


EDIT

One can ask "where does this method come from?" It follows from the hereafter observation. Take the generating functional of the theory: \begin{align} Z[J,\overline{\eta},\eta] =& \int \mathcal{D} g \mathcal{D} c \mathcal{D} \overline{c}\,e^{i S[g,c,\overline{c}]+i\int d^4x\,J^a_\mu g^{a,\mu}+i\int d^4x\,(\overline{\eta}{}^a c^a+\overline{c}{}^a \eta^a)} \end{align} Now, decompose $S$ into $S = S_\text{quad}+S_\text{int}$ where $S_\text{quad}$ contains all the quadratic terms in the fields, and $S_\text{int}$ all the interaction terms. Now pull the interaction action out of the generating functional, like this: \begin{align} Z[J,\overline{\eta},\eta] =& e^{iS_\text{int}\left[ \frac{-i\delta}{\delta J},\frac{i\delta}{\delta \overline{\eta}},\frac{-i\delta}{\delta \eta} \right]}\int \mathcal{D} A \mathcal{D} c \mathcal{D} \overline{c}\,e^{i S_\text{quad}[g,c,\overline{c}]+i\int d^4x\,J^a_\mu A^{a,\mu}+i\int d^4x\,(\overline{\eta}{}^a c^a+\overline{c}{}^a \eta^a)} \\ \stackrel{!}{=}& e^{iS_\text{int}\left[ \frac{-i\delta}{\delta J},\frac{i\delta}{\delta \overline{\eta}},\frac{-i\delta}{\delta \eta} \right]}e^{\frac{i}{2}\int d^4x \int d^4y\,J^a_\mu(x) G^{ab,\mu \nu}_{\text{gluon}} (x-y) J^b_\nu(y)}e^{i\int d^4x \int d^4y\,\overline{\eta}^a(x) G^{ab}_{\text{ghost}} (x-y) \eta^b(y)} \end{align} Now, you can easily use Fourier transform to have: \begin{align} Z[\tilde{J},\tilde{\overline{\eta}},\tilde{\eta}] =& e^{iS_\text{int}\left[ \frac{-i\delta}{\delta \tilde{J}},\frac{i\delta}{\delta \tilde{\overline{\eta}}},\frac{-i\delta}{\delta \tilde{\eta}} \right]}e^{\frac{i}{2}\int \frac{d^4 p}{(2\pi)^4} \int \frac{d^4 q}{(2\pi)^4}\,\tilde{J}^a_\mu(p) \tilde{G}^{ab,\mu \nu}_{\text{gluon}} (p-q) \tilde{J}^b_\nu(q)}e^{i\int \frac{d^4 p}{(2\pi)^4} \int \frac{d^4 q}{(2\pi)^4}\,\tilde{\overline{\eta}}^a(p) \tilde{G}^{ab}_{\text{ghost}} (p-q) \tilde{\eta}^b(q)} \end{align} Then, you can express this as a Weierstrass transform (see here, eq.(2.39)) as follows: \begin{align} Z[\tilde{J},\tilde{\overline{\eta}},\tilde{\eta}] =&\left. e^{\frac{i}{2}\int \frac{d^4 p}{(2\pi)^4} \int \frac{d^4 q}{(2\pi)^4}\,\frac{-i\delta}{\delta\tilde{J'}^{a}_\mu(p)} \tilde{G}^{ab,\mu \nu}_{\text{gluon}} (p-q) \frac{-i\delta}{\delta\tilde{J'}^b_\nu(q)}}e^{i\int \frac{d^4 p}{(2\pi)^4} \int \frac{d^4 q}{(2\pi)^4}\,\frac{i\delta}{\delta\tilde{\overline{\eta'}}^a(p)}\tilde{G}^{ab}_{\text{ghost}} (p-q) \frac{-i\delta}{\delta\tilde{\eta'}^b(q)}}e^{iS_\text{int}\left[ \tilde{J'},\tilde{\overline{\eta}'},\tilde{\eta'} \right]+\int d^4x (J'J+\overline{\eta}'\eta+\overline{\eta} \eta')} \right|_{J',\overline{\eta}',\eta'=0} \end{align} Where the primed fields are new dummy variables. For our purpose, we will make the following (schematical) expansion: \begin{equation} Z[\tilde{J},\tilde{\overline{\eta}},\tilde{\eta}] \sim \left. \left[ \int \frac{d^4p}{(2\pi)^4} \int \frac{d^4q}{(2\pi)^4} \frac{-i\delta}{\delta \tilde{J'}(p)}G(p-q) \frac{-i\delta}{\delta \tilde{J'}(q)} \right]^3 iS_\text{int,3G}\left[ \tilde{J'},\tilde{\overline{\eta}'},\tilde{\eta'} \right]^2+\cdots\right|_{J',\overline{\eta}',\eta'=0} \end{equation} Where the ellipsis contains all the other terms. This is a "sunset" bubble, constituted by three propagators and two vertices. To isolate one vertex, it suffices to take half of the derivative with respect to the sources, take only $S_\text{int, 3G}$ and not $S_\text{int, 3G}^2$, and ignore the propagators and the integrals. So for the 3-gluons vertex, we use the operator $\propto \frac{\delta^3}{\delta \tilde{J'}{}^a_\mu (p)\delta \tilde{J'}{}^b_\nu (q) \delta \tilde{J'}{}^c_\rho (k)}$. Or stated differently, upon changing the name of the dummy variable we differentiate over: \begin{equation} \Gamma^{abc}_{\mu \nu \rho}(p,q,k) \propto \frac{\delta^3 S_\text{int}[\tilde{g},\tilde{c},\tilde{\overline{c}}]}{\delta \tilde{g}{}^{a,\mu} (p)\delta \tilde{g}{}^{b,\nu} (q) \delta \tilde{g}{}^{c,\rho} (k)} \end{equation} This is indeed the formula I used before and it is a powerful one (when the interaction action doesn't have a too large power in the fields (otherwise the calculation becomes horrible)).


Note

I have $(2\pi)^4$ factors missing in my very first equation, but these disappear in the third one because we integrate out Dirac deltas in the impulsion representation.

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  • $\begingroup$ Thanks. But can you explain why you are taking the chain derivative and after that how you are plugging in the terms in the $S$ matrix element? $\endgroup$ Feb 27, 2023 at 19:23
  • $\begingroup$ @TanmoyPati This method comes from the path integral formalism. I've noticed that you did not include the interaction terms from the Faddeev-Popov ghosts (which are necessary) in your interaction Lagrangian. Since you are talking about the S matrix, are you familiar with the path integral formalism? $\endgroup$ Feb 28, 2023 at 6:14
  • $\begingroup$ I am not well efficient in path integral yet. But I have included the Faddeev Popov ghosts $\bar{c}$ and $c$ $\endgroup$ Feb 28, 2023 at 8:48
  • $\begingroup$ Whatever I know to calculate the amplitude in path integral, we have to couple the field with a current $J$ and variation with respect to $J(x)$, I don't know yet whether it is the only procedure. Here you are varying with respect to the gluons, So, if possible give some reference, which will be helpful for me. $\endgroup$ Feb 28, 2023 at 8:57
  • $\begingroup$ @TanmoyPati I will make an edit, you can find the procedure of deriving the gluons' vertices in any good book on QFT. (sorry for the FP ghosts, I was on my phone so the screen was more narrow and I didn't see them...) $\endgroup$ Feb 28, 2023 at 12:42

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