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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.....

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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 $\hbar$.

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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.

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