# How to draw one-loop corrections for a certain QFT theory

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).

• Do you mean to ask: what is the purpose of introducing the 1PI terms? What do they represent? Jun 16, 2020 at 21:09

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 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 $$\phi^2\chi$$.

• 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! Jun 16, 2020 at 22:13
• 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) Jun 16, 2020 at 22:48
• In momentum space, the external scalars contribute a factor of 1. So you can basically ignore external scalars in Feynman rules. Jun 16, 2020 at 23:07
• @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 $e$?
– Shen
Feb 28, 2021 at 12:23
• @Shen Why would you? Even the $Z^0$ boson is neutral and couples to charged fermions with a coupling constant which is different from $e$. Feb 28, 2021 at 12:42