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I would like to draw some Feynman diagrams using the feynmf package which contain a complex (triangular) vertex. Here is my attempt, which apparently reproduces the diagram which I want to achieve, however I have a few problems:

\begin{fmffile}{graf11}
{\begin{fmfgraph}(60,40)
\fmfleft{l}
\fmfright{r}
\fmfpoly{shaded}{z1,z2,z3}
\fmf{fermion,right=0.5}{l,z2}
\fmf{fermion,left=0.5}{l,z1}
\fmf{phantom}{z3,r}
\fmfdot{l}
\fmfdot{z1}
\fmfdot{z2}
\fmfdot{z3}
\end{fmfgraph}}
\end{fmffile}

My question is the following:

How can I draw only the triangular vertex without any propagation lines? I would like to avoid using phantom lines. It seems that drawing a standalone triangle (without any lines connected to it) with fmfpoly gives an error message.

  • Hello and welcome. Please add a minimal working example meta.tex.stackexchange.com/questions/228/… , with \documentclass ... \end{document}. – Bobyandbob Mar 15 '17 at 14:11
  • I reproduce your observation. However, you can make the propagation line very short by adding tension, e.g. \fmf{phantom,tension=99}{z3,r} . – marmot Mar 15 '17 at 14:53
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I'm not entirely sure what issue you are having. If it's just that you don't want to use a phantom line from v3 to r you could just use r in the fmfpoly directly and avoid need for an extra vertex.

The code is:

\begin{fmffile}{graf11}
{\begin{fmfgraph}(60,40)
\fmfleft{l}
\fmfright{r}
\fmfpoly{shaded}{z1,z2,r}
\fmf{fermion,right=0.5}{l,z2}
\fmf{fermion,left=0.5}{l,z1}
\fmfdot{l}
\fmfdot{z1}
\fmfdot{z2}
\fmfdot{r}
\end{fmfgraph}}

which gives the output Feynman Diagram

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