# Recipe for computing vertex factors in Feynman diagrams

I am currently studying quantum field theory from Srednicki. In class we have covered till chapter 14 and then skipped to IR divergences. So my knowledge of quantum field theory is limited to those sections. I just mentioned it in case it helps you guys to calibrate your responses to my level! Anyways, my question is related to determining vertex factors for any arbitrary interaction. We were introduced to vertex factors in chapter 9. For interacting theories such as: $$\mathcal{L}_I = \frac{g}{3!}\varphi^3$$ and $$\mathcal{L}_I = \frac{\lambda}{4!}\varphi^4$$ the vertex factors are $ig$ and $i\lambda$ respectively. Ignore the sign; I can never remember it. Also, for this discussion it probably won't be a big deal. Anyways, getting back to the main point, I only remember being told that the vertex factors are the ones given above but no one ever told me how they were obtained. Srednicki never formally defined them. I just assumed that you just read off the coefficient from the interacting part. Also, physically I have an intuition for the vertex factor. It represents the strength with which the interactions couple free fields. And when you compute Feynman diagrams more interactions means higher powers of this coupling parameter (or vertex factor).

In some later chapters Srednicki defined this vertex function $V_n(k_1,k_2,\dots,k_n)$; but he only formally talks about it in the context of vertex corrections. For the moment I'm not interested in corrections. I just want to know how you can compute this function for an arbitrary theory (say): $$\mathcal{L}_I = f_n(\varphi,\partial_\mu \varphi)$$ where the subscript $n$ the maximum number of fields and/or their derivatives defined in the function $f_n(\varphi,\partial_\mu \varphi)$. In general the vertex factor will not be just a constant like the two cases above. It will be a function of the incoming and outgoing four-momenta. What recipe should I follow to get that expression?

• Jun 8 '17 at 6:30

Here's a trick I always use: Suppose we have an interaction Lagrangian given by $$\mathcal{L}_{int} = \frac{\kappa}{3!}\phi^3.$$ The trick is to rewrite $$\mathrm{i}\times \text{interaction action}$$ in momentum space by introducing the Fourier decomposition of the fields as follows:
\begin{aligned} \mathrm{i}S_{int} &= \mathrm{i}\int\mathrm{d}^4x\, \mathcal{L}_{int} = \mathrm{i} \int\! \mathrm{d}^4x\, \frac{\kappa}{3!} [\phi(x)]^3 \\ &= \mathrm{i} \int\! \mathrm{d}^4x\, \frac{\kappa}{3!} \\ &\hspace{5mm}\times \int\! \frac{\mathrm{d}^4q_1}{(2\pi)^4}\, \mathrm{e}^{-\mathrm{i}q_1\cdot x} \widetilde{\phi}(q_1) \int\! \frac{\mathrm{d}^4q_2}{(2\pi)^4}\, \mathrm{e}^{-\mathrm{i}q_2\cdot x} \widetilde{\phi}(q_2) \int\! \frac{\mathrm{d}^4q_3}{(2\pi)^4}\, \mathrm{e}^{-\mathrm{i}q_3\cdot x} \widetilde{\phi}(q_3) \\ &=\int\!\frac{\mathrm{d}^4q_1\cdots\mathrm{d}^4q_3}{(2\pi\cdots2\pi)^4}\, (2\pi)^4 \delta^{(4)}(q_1+\cdots+q_3)\widetilde{\phi}(q_1)\widetilde{\phi}(q_2)\widetilde{\phi}(q_3)\times\frac{\mathrm{i} \kappa}{3!}. \end{aligned} Thus we can read off the vertex factor: $$\mathrm{i} \kappa / 3!$$ Don't forget to multiply by $3!$ which is the symmetry factor of the diagram. In the end you're simply left with $$\mathrm{i}\kappa.$$

Notice in the above the negative signs in all the exponentials, this means that we are taking all the momenta as incoming with respect to the vertex. That is, if you draw the diagram you must draw momentum arrows pointing towards the vertex: $q_1+q_2+q_3 = 0.$

Another example would be: $$\mathcal{L}_{int}:=g\bar{\psi}\gamma_{\mu}\gamma_{5}(\partial^\mu\phi)\psi,$$ where $g$ is a coupling constant and $\phi$ a scalar. The resulting vertex factor is (can you show this here?): $$V=g(\not{\!q}_2\gamma_5)_{jk},$$ where $q_2$ is the incoming momentum of $\phi$.

The $n$-point vertex is defined to be the 1PI $n$-point function with external lines amputated. So you just need to compute this correlation function and forget about external lines.

Let's see how this works for a $\lambda\phi^n/n!$ interaction at tree level. We get \begin{align*} \langle \prod_{i=1}^{n}\phi(k_i)\rangle=i\lambda \prod_{i=1}^{n}\langle \phi(k_i)\phi(-k_i)\rangle(2\pi)^d\delta^d(\sum_i k_i), \end{align*} and amputating the propagators indeed gives your answer $i\lambda$.

• Thanks for your response. I see how it will be constant for $\lambda \phi^n/n!$ in general. But what about $\partial_\mu \phi$ terms? Please also check my comment on the answer by @FredericBrünner Mar 21 '13 at 20:54

In general the vertex factor will not be just a constant like the two cases above. It will be a function of the incoming and outgoing four-momenta.

That is not true. In general, the vertex function is given by its momentum-dependent corrections to a constant vertex factor. For examples, see eq. 51.38 (Yukawa theory), 62.39 (QED) or 65.27 (scalar QED) in Srednicki.

• Check exercise number 10.4 from Srednicki. We have $\mathcal{L}_1 = \frac{1}{2}g\varphi \partial^\mu \varphi \partial_\mu \varphi$. The vertex factor turns out to be $\frac{1}{2}ig(k_1^2+k_2^2+k_3^2)$ Mar 21 '13 at 20:57
• PS: Having completed my 2nd quarter (Part 2) of my QFT class just now, I understand better what you mean by adding momentum-dependent vertex factors as corrections to the constant tree-level ones. Although our class ended at chapter 51. I may not understand the examples from chapters 62 and 65! Thanks for pointing them out to me. Mar 21 '13 at 20:59