2-loop correction to exact 3-point vertex in a complex scalar field theory with cubed interaction

Question

I am a graduate student with 1 quarter of relativistic QFT at the level of Srednicki (covered up to Chapter 30 this Fall). This question is not in any book that I know off and it wasn't assigned as homework but I tagged it under homework and exercises. The theory in question is similar to one in Srendicki but with a different interaction term (refer to Problem 9.3). Given this Lagrangian density: $$\mathcal{L} = -Z_\phi \partial^\mu \phi\partial_\mu\phi^\dagger -Z_m m^2\phi^\dagger\phi+\frac{1}{6}Z_g g(\phi^3 +(\phi^\dagger)^3)$$

where the scalar field $\phi$ is a complex field, are there any two-loop 1 particle irreducible corrections to the exact 3-point vertex function(s). I attempted to draw one but I am still unsure if it follows the Feynman rules for this theory.

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The reasoning for my attempt is the following:

1. Adopting the ("charged arrows") convention that lines are given arrows pointing away from $J$ sources and towards $J^\dagger$ sources, the vertices allowed by the theory are lines with all 3 arrows pointing either towards or away from the vertex. In case you don't know what I mean by the $J$'s, they come from adding a source term to the Lagrangian $J\phi^\dagger+J^\dagger\phi$ to take functional derivatives and then they are set to $J = J^\dagger = 0$. This is part of the typical argument used to obtain Feynman diagrams as done in Chapter 9 of Srednicki. I left out details regarding momentum labels here.
2. A correction to the 3-point vertex has to have 3 external lines.
3. I attempted to draw more "traditional" looking 2 loop vertex corrections taking diagrams from $\phi^3$ theory but it was not possible to assign arrows in a way that agrees with the vertices allowed by this theory. This is an example of such a diagram

After attempting to assign the charge arrows to many of these unsuccessfully, I went for a diagram where one of the external lines went over an internal line (but didn't touch). Since I have never seen one like this before and it doesn't seem like it breaks any rule, I thought this might be a valid two-loop correction.

Question: Does my attempted diagram (hand-drawn above) break any rules and is it generally a rule that an external line can go over an internal one?

Answer 1

1. Yes, Feynman diagrams do not have to be planar.

2. For $L=2$ loops and $E=3$ external legs, it follows

   • that a connected diagram has $L=I-(V-1)$, cf. e.g. eq. (8) in my Phys.SE answer here,

   • and that a 3-valent diagram has $3V=2I+E$,

   which leads to $V=5$ vertices and $I=6$ internal propagators.

3. Let us consider possible incoming 3-vertex 2-loop 1PI diagrams.

   • Such a diagram would consists of 3 incoming vertices $Y_i,Y_i,Y_i$ and 2 outgoing vertices $Y_o,Y_o$.

   • Each incoming external leg must be attached to a $Y_i$ vertex. Two external legs cannot be attached to the same $Y_i$ vertex since this would lead to a 1PR diagram. So each external leg is attached to a different $Y_i$.

   • Two vertices of same kind cannot be directly connected with a propagator as it would violate arrow direction.

   • The two internal lines of a $Y_i$ vertex cannot be attacted to the same $Y_o$ vertex since this would lead to a 1PR diagram. So each internal line of a $Y_i$ vertex is attacted to a different $Y_o$ vertex.

   It is not hard to see that OP's 1st diagram is the only possibility.

4. Similarly, OP's 2nd diagram is the only possible outgoing 3-vertex 2-loop 1PI diagram.