Precise definition of the vertex factor

Just a short question about the vertex factor in QFT. When I have an interaction Lagrangian

$$\mathcal{L}_{\mathrm{int}}=-\frac{\lambda}{3!}\phi^3$$

with a real scalar field $$\phi$$, is the vertex factor given by $$-i\lambda$$ or $$-i\frac{\lambda}{3!}$$?

Because as far as I learnt, the vertex factor is $$-i\frac{\lambda}{3!}$$ and $$3!$$ is the symmety factor of the diagram. But I saw in many books that they claim that $$-i\lambda$$ is the vertex factor of this interaction.....

2 Answers

It seems that OP is not questioning the standard convention to divide each term in the Lagrangian with its symmetry factor, e.g., $${\cal L}~=~-\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi -\frac{1}{2}m^2\phi^2 - \frac{\lambda}{3!}\phi^3.$$ Rather OP is assuming the above standard convention, and asks if the vertex factor$$^1$$ is $$-\frac{i}{\hbar}\lambda$$ or $$-\frac{i}{\hbar}\frac{\lambda}{3!}$$? The answer depends on context:

1. On one hand, viewing the 3-vertex as an amputated Feynman diagram (say, as the leading contribution to the 1PI 3-point vertex function), the vertex factor is $$-\frac{i}{\hbar}\lambda$$. An amputated Feynman diagram is typically stripped of the symmetry factor of the external legs because the legs are distinguishable -- they carry different momenta for starters.

2. On the other hand, e.g. in the source picture $$\phi\quad\longrightarrow\quad\frac{\hbar}{i}\frac{\delta}{\delta J} ,$$ where interaction terms $$-\frac{i}{\hbar}\frac{\lambda}{3!}\phi^3\quad\longrightarrow\quad-\frac{i}{\hbar}\frac{\lambda}{3!}\left(\frac{\hbar}{i}\frac{\delta}{\delta J}\right)^3$$ are differentiating propagator terms $$\frac{i}{2\hbar}J\Delta J$$ to build Feynman diagrams, the vertex factor is $$-\frac{i}{\hbar}\frac{\lambda}{3!}$$. For the source picture, see eq. (3) in my Phys.SE answer here.

References:

1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; first paragraph on p. 93.

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$$^1$$ We have restored the factors of $$\hbar$$.

It is conventional to write interactions normalized by the number of permutations of identical fields. So, there will be a $$\frac{1}{n!}$$ factor for each interaction with $$n$$ identical fields. This factor is then canceled by the $$n!$$ ways of permuting the $$n$$ identical lines coming out of the same internal vertex.

The diagram is therefore associated with just the prefactor, e.g. $$\lambda$$, from the interaction.

If the diagram presents a symmetry, you have also to divide by the geometrical symmetry factor.