3

I copied the code from Feynman Diagrams: Ugly u-channel, trying to produce a U-channel Feynman diagram. However, it produces a different, incorrect, Feynman diagram.

\documentclass[preview]{standalone}
\usepackage[compat=1.1.0]{tikz-feynman}

\begin{document}
\begin{tikzpicture}
  \begin{feynman}
    \diagram [vertical'=a to b] {
      i1 [particle=3]
      -- [fermion] a
      -- [draw=none] f1 [particle=1],
      a -- [photon, edge label'=(p)] b,
      i2 [particle=2]
      -- [anti fermion] b
      -- [draw=none] f2 [particle=4],
    };
    \diagram* {
      (a) -- [fermion] (f2),
      (b) -- [anti fermion] (f1),
    };
  \end{feynman}
\end{tikzpicture}
\end{document}

Here the correct Feynman diagram

correct Feynman diagram

My compiled image

wrong U-channel

How could it be that I can only produce the wrong Feynman diagram?

Jelmer Mulder
  • 31
  • Welcome to TeX.SX! Your chances for receiving an answer will increase if you explain how the graph should look like, and not just state that it's wrong. – gernot Feb 06 '21 at 23:02

1 Answers1

1

Here's a way by defining absolute vertices: somehow the solutions in the link seem to have some kind of problem.

result

\documentclass[10pt, border=10pt]{standalone}
\usepackage{tikz-feynman}

\begin{document}


\tikz{
  \begin{feynman}
    % left side: for i1 -- a -- b -- f1
    \vertex                         (i1) {(e^{-})};
    \vertex [below right=2cm of i1] (a);
    \vertex [below=1.5cm of a]      (b);
    \vertex [below left=1.5cm of b] (f1){(e^{+})};


% right side: i2(e+) and f2(e-)
\vertex [right=3cm of i1]       (i2) {\(e^{+}\)};
\vertex [right=3cm of f1]       (f2) {\(e^{-}\)};


\diagram* {
    % left side
    (i1) -- [fermion] (a),% e-
    (a) -- [boson, edge label'=\(p\)](b),
    (b) -- [fermion](f1),% e+

    % cross-overs at the right side
    (i2) -- [fermion] (b),
    (a) -- [fermion] (f2)
};




\end{feynman}
}


\end{document}
MS-SPO
  • 11,519