Feynman

A Feynman diagram showing sigma decay. I had used this diagram in One of my particle Physics homework. \begin{tikzpicture} \begin{feynman} \vertex (a); \vertex [below left=.4cm and 2cm of a] (b) {$s$}; \vertex [below right=.4cm and 2cm of a] (c){$u$}; \vertex [above=of a] (d); \vertex [below right=.4cm and 2cm of d] (e) {$d$}; \vertex [above right=.4cm and 2cm of d] (f) {$u$}; \vertex [below=0.2cm of b] (s1) {$s$}; \vertex [below=0.2cm of c] (s2) {$s$}; \vertex [below=0.45cm of b] (s3){$d$}; \vertex [below=0.45cm of c] (s4){$d$}; \diagram*{ (b) --[fermion] (a) --[fermion] (c), (a) --[boson,edge label'={\small $W^{-}$}] (d), (d) --[fermion] (e), (d) -- [anti fermion] (f), (s1) -- [fermion] (s2); (s3) -- [fermion] (s4); }; \end{feynman} \end{tikzpicture}

A feynman diagram for neutrinoless double beta decay $0\nu\beta\beta$. Refer to this better version \begin{tikzpicture} \begin{feynman} \vertex (b); \vertex [below=of b] (c); \vertex [below left=1cm and 1.4cm of c] (d); \vertex [above left=.8cm and 1.4cm of b] (a); \vertex [left=of a] (i1) {\tiny $d$}; \vertex [left=of d] (i2) {\tiny $d$}; \vertex [right = 2cm of b] (f2) {\tiny $e^{-}$}; \vertex [right = 2cm of c] (f3) {\tiny $e^{-}$}; \vertex [below = 2cm of f3] (f4) {\tiny $u$}; \vertex [above = 2cm of f2] (f1) {\tiny $u$}; \vertex [above=0.25cm of i1] (f6) {\tiny $d$}; % d quark outgoing \vertex [above=0.25cm of f1] (i3) {\tiny $d$}; % d quark ingoing \vertex [above=0.25cm of i3] (f7) {\tiny $u$}; % u quark outgoing \vertex [above=0.25cm of f6] (i4) {\tiny $u$}; % u quark ingoing % copy quarks for bottom \vertex [below=0.25cm of i2] (f8) {\tiny $d$}; % d quark outgoing \vertex [below=0.25cm of f4] (i5) {\tiny $d$}; % d quark ingoing \vertex [below=0.25cm of i5] (f9) {\tiny $u$}; % u quark outgoing \vertex [below=0.25cm of f8] (i6) {\tiny $u$}; % u quark ingoing \newcommand\tmpda{0.9cm} \newcommand\tmpdb{-1.7cm} \diagram* { (a) -- [boson, edge label'= {\tiny $W^{-}$}] (b) -- [anti majorana, insertion=0.5, edge label' = {\tiny $\nu_{\scriptsize M}$} ] (c) -- [boson, edge label'={\tiny $W^{-}$}] (d), (i1) -- [with arrow=\tmpda] (a), (i2) -- [with arrow=\tmpda] (d), (a) -- [with arrow=\tmpdb] (f1), (b) -- [fermion] (f2), (c) -- [fermion] (f3), (d) -- [with arrow=\tmpdb] (f4), (f6) -- [with arrow=\tmpda, with arrow=\tmpdb, out=0, in=200] (i3), (i4) -- [with arrow=\tmpda, with arrow=\tmpdb, out=0, in=200] (f7), (f8) -- [with arrow=\tmpda, with arrow=\tmpdb, out=0, in=160] (i5), (i6) -- [with arrow=\tmpda, with arrow=\tmpdb, out=0, in=160] (f9), }; \draw [decoration = {brace} , decorate] (i1.south west) -- (i4.north west) node [pos = 0.5 , left = 0.065cm] {\small $n$}; \draw [decoration = {brace} , decorate] (f7.north east) -- (f1.south east) node [pos = 0.5 , right = 0.125cm] {\small p}; %J\draw [decoration = {brace} , decorate] (i6.south west) -- (i2.north west) node [pos = 0.5 , left = 0.125cm] {\small n}; %\draw [decoration = {brace} , decorate] (f4.north east) -- (f9.south east) node [pos = 0.5 , right = 0.125cm] {\small p}; \end{feynman} \end{tikzpicture}

A feynman diagram for neutrinoless double beta decay $0\nu\beta\beta$. Time obviously points upwards rather than sideways from left to right. As my professor said, “I don’t understand why people use left to right to indicate time. Afterall Feynman himself used bottom to top.” \begin{tikzpicture} \begin{feynman} \vertex (a); \vertex [below left=2cm and 0.4cm of a] (b); \vertex [above left=2cm and 0.4cm of a] (c); \vertex [above right=of a] (d); \vertex [right= 0.70cm of d] (j); \vertex [above left=1cm and 0.4cm of d] (e); \vertex [above right=1cm and 0.4cm of d] (f); \diagram*{ (b)--[fermion,edge label={$n^0$}] (a) --[fermion,edge label={$p^{+}$}] (c), (a) --[boson,edge label'={\small $W^{-}$}] (d), (d) --[fermion,edge label={$e^{-}$}] (e), (j) --[fermion,edge label'={\tiny $\bar{\nu}$}] (d) }; \vertex [right=3.5cm of a] (aa); \vertex [below right=2cm and 0.4cm of aa] (ab); \vertex [above right=2cm and 0.4cm of aa] (ac); \vertex [above left=of aa] (ad); \vertex [above left=1cm and 0.4cm of ad] (ae); \vertex [above right=1cm and 0.4cm of ad] (af); \diagram*{ (ab)--[fermion,edge label'={$n^0$}] (aa) --[fermion, edge label'={$p^{+}$}] (ac), (aa) --[boson,edge label={\small $W^{-}$}] (ad), (ad) --[fermion,edge label'={ $e^{-}$}] (af), (j) --[fermion,edge label={\tiny $\bar{\nu}$}] (ad), (j) --[insertion=0.0] (d) }; \end{feynman} \end{tikzpicture}

A feynman diagram for two neutrino double beta decay $2\nu\beta\beta$. Time obviously points upwards rather than sideways from left to right. As my professor said, “I don’t understand why people use left to right to indicate time. Afterall Feynman himself used bottom to top.” \begin{tikzpicture} \begin{feynman} \vertex (a); \vertex [below left=2cm and 0.4cm of a] (b); \vertex [above left=2cm and 0.4cm of a] (c); \vertex [above right=of a] (d); \vertex [above left=1cm and 0.4cm of d] (e); \vertex [above right=1cm and 0.4cm of d] (f); \diagram*{ (b)--[fermion,edge label={$n^0$}] (a) --[fermion,edge label={$p^{+}$}] (c), (a) --[boson,edge label'={\small $W^{-}$}] (d), (d) --[fermion,edge label={$e^{-}$}] (e), (f) --[fermion,edge label'={\tiny $\bar{\nu}$}] (d) }; \vertex [right=3.5cm of a] (aa); \vertex [below right=2cm and 0.4cm of aa] (ab); \vertex [above right=2cm and 0.4cm of aa] (ac); \vertex [above left=of aa] (ad); \vertex [above left=1cm and 0.4cm of ad] (ae); \vertex [above right=1cm and 0.4cm of ad] (af); \diagram*{ (ab)--[fermion,edge label'={$n^0$}] (aa) --[fermion, edge label'={$p^{+}$}] (ac), (aa) --[boson,edge label={\small $W^{-}$}] (ad), (ae) --[fermion,edge label={\tiny $\bar{\nu}$}] (ad), (ad) --[fermion,edge label'={ $e^{-}$}] (af) }; \end{feynman} \end{tikzpicture}