Respuesta1
Aquí hay una colección de diagramas para la teoría de campos libres, proporcionados como phinman.sty
(código al final de esta respuesta). Por motivos de espaciado, hay dos versiones de cada comando, una con etiquetas y otra sin ellas.
El código para la demostración anterior:
\documentclass{article}
\usepackage{phinman} % pronounce like Feynman
\usepackage{amsmath}
\usepackage{lipsum}
\begin{document}
\begin{center}
\begin{tabular}{lc@{\quad}lc}
\verb"\PM{-}" & \PM{-} & \verb"\PMl{-}{x}{y}" & \PMl{-}{x}{y} \\
\verb"\PM{x}" & \PM{x} & \verb"\PMl{x}{x}" & \PMl{x}{x} \\
\verb"\PM{||}" & \PM{||} & \verb"\PMl{||}{u}{v}{x}{y}" & \PMl{||}{u}{v}{x}{y} \\
\verb"\PM{=}" & \PM{=} & \verb"\PMl{=}{u}{v}{x}{y}" & \PMl{=}{u}{v}{x}{y} \\
\verb"\PM{X}" & \PM{X} & \verb"\PMl{X}{u}{v}{x}{y}" & \PMl{X}{u}{v}{x}{y} \\
\verb"\PM{--}" & \PM{--} & \verb"\PMl{--}{x}{y}" & \PMl{--}{x}{y} \\
\verb"\PM{-o-}" & \PM{-o-} & \verb"\PMl{-o-}{x}{y}{z}" & \PMl{-o-}{x}{y}{z} \\
\verb"\PM{8}" & \PM{8} & \verb"\PMl{8}{x}" & \PMl{8}{x} \\
\verb"\PM{88}" & \PM{88} & \verb"\PMl{88}{x}{y}" & \PMl{88}{x}{y} \\
\verb"\PM{ooo}" & \PM{ooo} & \verb"\PMl{ooo}{x}{y}" & \PMl{ooo}{x}{y} \\
\verb"\PM{-8-}" & \PM{-8-} & \verb"\PMl{-8-}{x}{y}{z}" & \PMl{-8-}{x}{y}{z} \\
\verb"\PM{-88-}" & \PM{-88-} & \verb"\PMl{-88-}{x}{u}{v}{y}" & \PMl{-88-}{x}{u}{v}{y} \\
\verb"\PM{-ooo-}"& \PM{-ooo-}& \verb"\PMl{-ooo-}{x}{u}{v}{y}"& \PMl{-ooo-}{x}{u}{v}{y}\\
\verb"\PM{-o8-}" & \PM{-o8-} & \verb"\PMl{-o8-}{x}{u}{v}{y}" & \PMl{-o8-}{x}{u}{v}{y}
\end{tabular}
\end{center}
\clearpage
\lipsum[1]
\begin{align*}
\PMl{-}{x}{y} & = i\Delta(x,y)\\
\PMl{x}{y} & = -i\lambda\\
\langle\phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\rangle_0
&= \PMl{||}{x_1}{x_2}{x_3}{x_4} + \PMl{=}{x_1}{x_2}{x_3}{x_4}
+ \PMl{X}{x_1}{x_2}{x_3}{x_4}\\
\PMl{-o-}{x_1}{x_2}{y}&=\frac{-i\lambda}2\int\cdots\\
\PMl{-8-}{x_1}{y}{x_2}&=\frac{-i\lambda}6i\Delta(x_1,x_2)\int\cdots\\
\PMl{8}{y} &=\frac{-i\lambda}6\int\cdots\\
\text{denominator} &=1+\PM{8}+\PM{88}+\PM{ooo}+\cdots\\
\text{numerator} &=\PM{--}+\PM{-8-}+\PM{-88-}+\PM{-ooo-}+\cdots\\
&\quad+\PM{-o-}+\PM{-o8-}+\cdots
\end{align*}
\lipsum[2]
\end{document}
El código del paquete phinman.sty
(se pronuncia como Feynman):
\NeedsTeXFormat{LaTeX2e}
\ProvidesPackage{phinman}[2017/04/14 Diagrams for the free field theory]
\RequirePackage{tikz}
\newcommand\PMset{\pgfqkeys{/PM}}
\newcommand\PMldef[1]{\expandafter\def\csname PMl:#1\endcsname}
\newcommand\PMl[1]{\csname PMl:#1\endcsname}
\newcommand\PMdef[1]{\expandafter\def\csname PM:#1\endcsname}
\newcommand\PM[1]{\csname PM:#1\endcsname}
\newlength\PMu
\PMu=2ex
\PMset
{dot/.style={circle,fill,draw,inner sep=0pt,outer sep=0pt,minimum size=2.5pt},
line/.style={line width=0.6pt},
ghostline/.style={line width=2pt,color=white}
}
\newcommand\PMdot{node[/PM/dot]{}}
\newcommand\PMdotx[1]{node[/PM/dot,label={#1}]{}}
\newcommand\PMdota[1]{\PMdotx{[yshift=-0.5ex]above:{$#1$}}}
\newcommand\PMdotb[1]{\PMdotx{[yshift=0.5ex]below:{$#1$}}}
\newcommand\PMdotl[1]{\PMdotx{[xshift=0.5ex]left:{$#1$}}}
\newcommand\PMdotr[1]{\PMdotx{[xshift=-0.5ex]right:{$#1$}}}
\PMdef{-}%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdot -- (2\PMu,0)\PMdot;%
}
\PMldef{-}#1#2%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdotb{#1} -- (2\PMu,0)\PMdotb{#2};%
}
\PMdef{--}%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdot -- (4\PMu,0)\PMdot;%
}
\PMldef{--}#1#2%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdotb{#1} -- (4\PMu,0)\PMdotb{#2};%
}
\PMdef{x}%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdot
(\PMu,\PMu) -- (-\PMu,-\PMu)
(\PMu,-\PMu) -- (-\PMu,\PMu);%
}
\PMldef{x}#1%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdotx{below:{$#1$}}
(\PMu,\PMu) -- (-\PMu,-\PMu)
(\PMu,-\PMu) -- (-\PMu,\PMu);%
}
\PMdef{||}%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,\PMu) \PMdot -- (0,-\PMu) \PMdot
(2\PMu,\PMu) \PMdot -- (2\PMu,-\PMu)\PMdot;%
}
\PMldef{||}#1#2#3#4%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,\PMu) \PMdotl{#3} -- (0,-\PMu) \PMdotl{#4}
(2\PMu,\PMu) \PMdotr{#1} -- (2\PMu,-\PMu)\PMdotr{#2};%
}
\PMdef{=}%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,\PMu) \PMdot -- (2\PMu,\PMu) \PMdot
(0,-\PMu) \PMdot -- (2\PMu,-\PMu)\PMdot;%
}
\PMldef{=}#1#2#3#4%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,\PMu) \PMdotl{#3} -- (2\PMu,\PMu) \PMdotr{#1}
(0,-\PMu) \PMdotl{#4} -- (2\PMu,-\PMu)\PMdotr{#2};%
}
\PMdef{X}%
{\tikz[baseline={(0,-0.5ex)}]%
{\draw[/PM/line] (0,\PMu) \PMdot -- (2\PMu,-\PMu)\PMdot;
\draw[/PM/ghostline] (0,-\PMu)-- (2\PMu,\PMu);
\draw[/PM/line](0,-\PMu)\PMdot -- (2\PMu,\PMu) \PMdot;
}%
}
\PMldef{X}#1#2#3#4%
{\tikz[baseline={(0,-0.5ex)}]%
{\draw[/PM/line] (0,\PMu) \PMdotl{#3} -- (2\PMu,-\PMu)\PMdotr{#2};
\draw[/PM/ghostline] (0,-\PMu)-- (2\PMu,\PMu);
\draw[/PM/line](0,-\PMu)\PMdotl{#4}-- (2\PMu,\PMu) \PMdotr{#1};
}%
}
\PMdef{-o-}%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdot -- ++(2\PMu,0)\PMdot
arc(-90:270:0.6\PMu) -- ++(2\PMu,0)\PMdot;%
}
\PMldef{-o-}#1#2#3%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdotb{#1} -- ++(2\PMu,0)\PMdotb{#2}
arc(-90:270:0.6\PMu) -- ++(2\PMu,0)\PMdotb{#3};%
}
\PMdef{-8-}%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdot -- ++(4\PMu,0)\PMdot
(2\PMu,0.8\PMu)\PMdot arc (0:360:0.6\PMu)
arc (180:540:0.6\PMu);%
}
\PMldef{-8-}#1#2#3%
{\tikz[baseline={(0,0.5ex)}]\draw[/PM/line]
(0,0) \PMdotb{#1} -- ++(4\PMu,0)\PMdotb{#3}
(2\PMu,0.8\PMu)\PMdotx{above:{$#2$}} arc (0:360:0.6\PMu)
arc (180:540:0.6\PMu);%
}
\PMdef{-ooo-}%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdot -- ++(4\PMu,0)\PMdot
(1.3\PMu,0.8\PMu)\PMdot arc (0:360:0.6\PMu)
arc (180:360:0.6\PMu) \PMdot
arc (180:540:0.6\PMu) arc (0:180:0.6\PMu);%
}
\PMldef{-ooo-}#1#2#3#4%
{\tikz[baseline={(0,0.5ex)}]\draw[/PM/line]
(0,0) \PMdotb{#1} -- ++(4\PMu,0)\PMdotb{#4}
(1.3\PMu,0.8\PMu)\PMdotx{above:{$#2$}} arc (0:360:0.6\PMu)
arc (180:360:0.6\PMu) \PMdotx{above:{$#3$}}
arc (180:540:0.6\PMu) arc (0:180:0.6\PMu);%
}
\PMdef{8}%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdot arc (0:360:0.6\PMu)
arc (180:540:0.6\PMu);%
}
\PMldef{8}#1%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdotx{below:{$#1$}} arc (0:360:0.6\PMu)
arc (180:540:0.6\PMu);%
}
\PMdef{88}%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,-0.7\PMu) \PMdot arc (0:360:0.6\PMu) arc (180:540:0.6\PMu)
(0,0.7\PMu) \PMdot arc (0:360:0.6\PMu) arc (180:540:0.6\PMu);%
}
\PMldef{88}#1#2%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,-0.7\PMu) \PMdotx{below:{$#2$}} arc (0:360:0.6\PMu) arc (180:540:0.6\PMu)
(0,0.7\PMu) \PMdotx{above:{$#1$}} arc (0:360:0.6\PMu) arc (180:540:0.6\PMu);%
}
\PMdef{ooo}%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdot arc (0:360:0.6\PMu)
arc (180:360:0.6\PMu) \PMdot
arc (180:540:0.6\PMu) arc (0:180:0.6\PMu);%
}
\PMldef{ooo}#1#2%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,0) \PMdotx{below:{$#1$}} arc (0:360:0.6\PMu)
arc (180:360:0.6\PMu) \PMdotx{below:{$#2$}}
arc (180:540:0.6\PMu) arc (0:180:0.6\PMu);%
}
\PMdef{-88-}%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,-0.8\PMu) \PMdot arc (0:360:0.6\PMu) arc (180:540:0.6\PMu)
(-2\PMu,0) \PMdot -- (2\PMu,0)\PMdot
(0,0.8\PMu) \PMdot arc (0:360:0.6\PMu) arc (180:540:0.6\PMu);%
}
\PMldef{-88-}#1#2#3#4%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,-0.8\PMu) \PMdotx{below:{$#3$}} arc (0:360:0.6\PMu) arc (180:540:0.6\PMu)
(-2\PMu,0) \PMdotb{#1} -- (2\PMu,0)\PMdotb{#4}
(0,0.8\PMu) \PMdotx{above:{$#2$}} arc (0:360:0.6\PMu) arc (180:540:0.6\PMu);%
}
\PMdef{-o8-}%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,-0.8\PMu) \PMdot arc (0:360:0.6\PMu) arc (180:540:0.6\PMu)
(-2\PMu,0) \PMdot -- (0,0) \PMdot
arc (-90:270:0.6\PMu) -- (2\PMu,0)\PMdot;%
}
\PMldef{-o8-}#1#2#3#4%
{\tikz[baseline={(0,-0.5ex)}]\draw[/PM/line]
(0,-1.5\PMu) \PMdotx{below:{$#3$}} arc (0:360:0.6\PMu) arc (180:540:0.6\PMu)
(-2\PMu,0) \PMdotb{#1} -- (0,0) \PMdotb{#2}
arc (-90:270:0.6\PMu) -- (2\PMu,0)\PMdotb{#4};%
}
\endinput
Respuesta2
Puedes dibujarlo con Tikz. A continuación he creado un par de comandos para los símbolos en tu dibujo (probablemente puedas encontrar mejores nombres para ellos). Tuve que ajustar la altura 2pt
manualmente para alinear las líneas con los +
carteles. Las lipsum
s son sólo para ver el espaciado vertical. Puedes ajustar la longitud de las líneas y el tamaño de los círculos.
\documentclass{article}
\usepackage{amsmath}
\usepackage{lipsum}
\usepackage{tikz}
\usetikzlibrary{calc,arrows}
\newcommand\SajadLineLength{15mm}
\newcommand\SajadCircleRadii{3mm}
\newcommand\SajadLine[2]{\tikz[baseline=-2pt]{%
\draw[*-*](0,0) -- (\SajadLineLength,0)node[pos=0,below]{$#1$}node[pos=1,below]{$#2$};
}}
\newcommand\SajadLineCircleMid[2]{\tikz[baseline=-2pt]{%
\draw[*-*](0,0) -- (\SajadLineLength,0)node[pos=0,below]{$#1$}node[pos=1,below]{$#2$};
\draw (0.5*\SajadLineLength,0) circle (\SajadCircleRadii);
}}
\newcommand\SajadLineCircle[2]{\tikz[baseline=-2pt]{%
\draw[*-*](0,0) -- (\SajadLineLength,0)node[pos=0,below]{$#1$}node[pos=1,below]{$#2$};
\draw (0.5*\SajadLineLength,\SajadCircleRadii) circle (\SajadCircleRadii);
}}
\newcommand\SajadLineDoubleCircle[2]{\tikz[baseline=-2pt]{%
\draw[*-*](0,0) -- (\SajadLineLength,0)node[pos=0,below]{$#1$}node[pos=1,below]{$#2$};
\draw (0.5*\SajadLineLength,\SajadCircleRadii) circle (\SajadCircleRadii);
\draw (0.5*\SajadLineLength,3*\SajadCircleRadii) circle (\SajadCircleRadii);
}}
\begin{document}
\lipsum[1]
\begin{displaymath}
A=
\SajadLine{x_1}{x_2}
+\SajadLineCircle{x_1}{x_2}
+\SajadLineCircleMid{x_1}{x_2}
+\SajadLineDoubleCircle{x_1}{x_2}
\end{displaymath}
\lipsum[2]
\end{document}