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Consider that I'm given the following Lagrangean:

$$L=L_{QED}+\frac{1}{2}\partial_\mu\phi\partial^\mu\phi+\partial_\mu\chi\partial^\mu\chi-\frac{1}{2}m_\phi^2\phi^2-\frac{1}{2}m_\chi^2\chi^2-\frac{1}{2}\mu_1\phi^2\chi-\frac{1}{2}\mu_2\chi^2\phi-g\bar{\psi}\psi\chi$$

where $\phi$ and $\chi$ are neutral scalar fields and $\psi$ is the electron field. From this Lagrangean I extract new Feynman rules besides QED's, for instance, the propagator for the scalar fields, the vertices of electrons and $\chi$ field, vertices between the scalar fields, etc.

Within QED, in particular, the one-loop correction to the vertex is done by introducing a photon propagator "connecting" the fermionic lines. Furthermore, the one-loop correction for the electron propagator introduces the electron's self energy and for the photon propagator, the vacuum polarization.

My question is, why are these the one-loop 1PI corrections to QED? How do I know that for the vertex, I need to connect those two electron lines with a photon? The reason I ask this is because I'm trying to draw the one-loop corrections for the scalar self energy $\chi$ and for the new vertices (so I can discuss their superficial degree of divergence after).

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  • $\begingroup$ Do you mean to ask: what is the purpose of introducing the 1PI terms? What do they represent? $\endgroup$
    – StudyStudy
    Jun 16 '20 at 21:09
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To construct Feynman diagrams you have to look up at the interaction terms in your lagrangian. The fact that in QED you have a $\gamma e^-e^+$ vertex comes directly from the interaction term between the photon field and the fermion field in the QED lagrangian $$e\bar\psi\not A\psi\to-ie\gamma^\mu$$

With this interaction vertex you can build up only certain diagrams. For example the four photon interaction diagram in QED needs to be done using a fermion loop since there's not four photon interaction term in the lagrangian.

Now, in your lagrangian you have three more interaction vertices between the fields: a vertex with two $\phi$ and a $\chi$ given by the term $$\mu_1\phi^2\chi$$ for which the Feynman rule gives $-i\mu_1$, a vertex with two $\chi$ and one $\phi$ $$\mu_2\chi^2\phi$$ for which the Feynman rule gives $-i\mu_2$, and a vertex with two fermions and a $\chi$ $$g\bar\psi\psi\chi$$ for which the Feynman rule gives $-ig$. For example this last vertex gives another one loop contribution to the fermion propagator which is identical to the fermion self energy diagram but with the photon replaced by a $\chi$.

If you're searching for the one loop corrections to the scalar propagator of the $\chi$ you'll have three diagrams enter image description here

The first is given by the interaction term $\bar\psi\psi\chi$, the second by the interaction $\chi^2\phi$ (sorry if both the $\chi$ and the $\phi$ are given by dashed lines), and the third is the remaining term $\phi^2\chi$.

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  • $\begingroup$ Thank you very much for your answer! I initially also thought that any new correction could only appear according to the existing interactions and I was relieved to see I got the same corrections to the scalar propagator of $\chi$ as you did! $\endgroup$
    – RicardoP
    Jun 16 '20 at 22:13
  • $\begingroup$ An extra question: While fermions contribute with spinors to the amplitude of any diagram, do external lines of scalar fields contribute with anything? (i'm assuming they don't) $\endgroup$
    – RicardoP
    Jun 16 '20 at 22:48
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    $\begingroup$ In momentum space, the external scalars contribute a factor of 1. So you can basically ignore external scalars in Feynman rules. $\endgroup$
    – GrassyNol
    Jun 16 '20 at 23:07
  • $\begingroup$ @Davide Morgante - For the vertex $g\overline{\psi}\psi \chi$, is the coupling constant $g$ the same as $e$ (since $\psi$ is the electron field and $\chi$ is neutral)? Can we just replace $g$ by $e$? $\endgroup$
    – Shen
    Feb 28 at 12:23
  • $\begingroup$ @Shen Why would you? Even the $Z^0$ boson is neutral and couples to charged fermions with a coupling constant which is different from $e$. $\endgroup$ Feb 28 at 12:42

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