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A real scalar field has Lagrangian $L=L_0+L_1$, where the free part $$L_0=-\frac{1}{2}(\partial\phi)^2-\frac{1}{2}m^2\phi^2$$ and the interaction term $$L_1 = + \frac{1}{2}g\phi^2.$$

I have two separate questions about this.

  1. How can an interaction term can be quadratic in the field? I thought this meant the theory was 'free'.
  2. What are the momentum space formulas for the propagator and vertex for this theory?
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There is no absolute rule to be followed in order to determine if something is an "interaction" or not. You can interpret it in different ways, do field redefinitions, etc. At the end of the day, if you are consistent with your computations, you should get the same answer that nature gives when an experiment is done.

So, you might as well treat the term quadratic in $\phi$ as an interaction. For a 2-point function, draw the first few diagrams in increasing powers of the coupling in the quadratic interaction. It will just be a line connecting the two ends of a 2-point function, with dots in between, each dot representing the quadratic interaction. We never really see this kind of Feynman diagram because it's trivial to simplify: just sum all the diagrams in series of the quadratic interaction coupling. Your answer should match the propagator for the theory with the quadratic part included in the free theory. It's just a different way of doing perturbation theory, but of course, physical results don't change by how you wish to perform your perturbations.

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Promoted to an answer from a comment by flippiefanus:

The quadratic "interaction" term is just a mass term. In your case it will combine with the existing mass term and they'll act together as a modified mass term. So, yes you are right, the theory is free.

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