TITLE: Feynman Diagrams for a quadratic interaction term
QUESTION [2 upvotes]: Consider the Lagrangian
$$\mathcal{L} = \dfrac{1}{2} (\partial_{\mu}\phi_{1})^2 + \dfrac{1}{2}(\partial_{\mu}\chi)^2 - \dfrac{M_1^2}{2}\phi_{1}^2 -\dfrac{M^2_\chi}{2} \chi^2 - \dfrac{\mu_\chi}{2} \phi_1\chi^2 $$
where $\phi_1$ and $\chi$ are real scalar fields. I know that for the above Lagrangian, using only the interaction term $\mathcal{L}_{int} =-\dfrac{M_1^2}{2}\phi_{1}^2$, we recover the Feynman propagator for the 2 pt function $<0 T\lbrace \phi_1(x)\phi_1(y)\rbrace0>$ after summing up all the degenerate Feynman diagrams i.e we get:
$$i\Pi=\dfrac{i}{p^2-M_1^2 +i\epsilon }$$
Now consider the following Lagrangian with an extra massive real scalar field $\phi_2$ and a quadratic mixing term $- M_{12}^2\phi_1\phi_2$,
$$\mathcal{L} = \dfrac{1}{2} (\partial_{\mu}\phi_{1})^2 + \dfrac{1}{2}(\partial_{\mu}\phi_{2})^2 + \dfrac{1}{2}(\partial_{\mu}\chi)^2 - \dfrac{M_1^2}{2}\phi_{1}^2 - \dfrac{M_2^2}{2}\phi_{2}^2 - M_{12}^2\phi_1\phi_2 - \dfrac{M^2_\chi}{2} \chi^2 - \dfrac{\mu_\chi}{2} \phi_1\chi^2 $$
In this case, how do I draw and evaluate the Feynman diagrams that contribute to the 2 point function from using only the quadratic term  $- M_{12}^2\phi_1\phi_2$ as an interaction up to order $M^6_{12}$?
How does this differ from the previous Lagrangian and how the propagator will be different now?

REPLY [1 votes]: The Feynman diagram for a term of the form $-M_{12}^2\phi_1\phi_2$ would be a single line coming into a dot and a single line coming out. If the line in is $\phi_1$ the line out is $\phi_2$, and vice-versa. Connecting these up to form higher order diagrams is left as an exercise for the reader.
{"set_name": "stack_exchange", "score": 2, "question_id": 397228}