In QFT why does the degree of the interaction terms in Lagrangian start from 3? I'm new to QFT so it's not obvious to me why there is no quadratic interaction terms in Lagrangians.
For example, the Lagrangian for a real scalar field is
$$L=\frac{1}{2}\partial_\mu \phi \partial^\mu \phi-\frac{1}{2}m^2\phi^2-\sum_{n\geq 3}\frac{\lambda_n}{n!}\phi^n.$$
What's the reason that we can't add terms like $g\phi^2$ to the free field Lagrangian?
 A: 
...why does the degree of the interaction terms in Lagrangian start from 3?


...What's the reason that we can't add terms like $g\phi^2$ to the free field Lagrangian?

You can add such a $g\phi^2$ term if you want, but it's not usually called an "interaction" term, since if $g$ is constant you can just combine it with the mass term to have another free Lagrangian with a different mass.
If the $g$ parameter is not constant then such a term could represent an external potential (still not an "interaction" though). However, we often don't add an external potential term in elementary particle theory because there are usually no real external potentials in this context. In condensed matter field theory, external potential potentials are more common.
A: Such a term can be absorbed into a redefinition of the propagator.
A: Given that an interaction of the $n^{th}$ order implies the existence of an $n$ pointed vertex.
ex. $L_1 = g \phi^3$ in a real scalar field theory gives rise to a $3$ pointed vertex.
If the interaction term had form $$L_1 = g\phi^2$$ this would just be a $2$ pointed vertex. If we allow conservation laws in this theory then a particle going into the vertex with energy and momentum $(p_i, E_i)$ will have to be equivalent to the outgoing one, making $(p_f, E_f) = (p_i, E_i)$. a.k.a, it's the same particle and nothing changed.
It is interesting to see that given a free Lagrangian
$$L_0 = \frac{1}{2}\partial^\mu\phi\partial_\mu\phi + \frac{1}{2}m^2\phi^2,$$ by applying the previous interpretation here, $m$ could be seen as the coupling strength between a free non-interacting particle and, nothing.
A: Contrary to what the other answer's claim, we do add these $g\phi^2$ terms as interaction term.
Check any QFT text book, specifically the section on mass renormalization and field renormalization counter terms.
