关于 TikZ 箭头（大小）用于标记/指向方程式的问题

Question

我正在尝试使用箭头来标记一组方程式。这是我目前所得到的。

 \usepackage{tikz} % needed for arrows explaining equations
 \usetikzlibrary{tikzmark,positioning}
 \tikzstyle{every picture}+=[remember picture]
 \everymath{\displaystyle}
 %% the above lines go into the preamble
 \tikzstyle{na} = [baseline=-.5ex]
 \begin{align*}
 \begin{array}{cll}
           & \tikz[baseline]{\node[anchor=base](eq1){}} \delta\lambda = -\dot{\varphi}\epsilon & +\epsilon\tfrac{\partial}{\partial p}S = \epsilon(\dot{\varphi}-p)\\
 \text{Text1}\tikz[na]\node[coordinate](eom1){};   &  &\\ 
             & \tikz[baseline]{\node[anchor=base](eq2){}} \delta p = i\epsilon(x\dot{\lambda} + (1-x)2i\dot{\pi}) & -\epsilon\tfrac{\partial}{\partial\lambda}S = \epsilon\dot{\pi}\\
         & &   \\
           &                                        & +\tfrac{i}{2}\epsilon\tfrac{\partial}{\partial \pi}S = -\tfrac{i}{2}\epsilon\dot{\lambda}\\
  \text{Text2}\tikz[na]\node[coordinate](eom2){};  &                                     &\\ 
            &\tikz[baseline]{\node[anchor=base](eq3){}} \delta\pi = -\tfrac{i}{2}(-\dot{\varphi}\epsilon) & -\tfrac{i}{2}\epsilon\tfrac{\partial}{\partial p}S = -\tfrac{i}{2}(\dot{\varphi}-p)\epsilon
 \end{array}
 \begin{tikzpicture}[overlay]
     \path[->]<1-> (eom1) edge [right] (eq1);
     \path[->]<1-> (eom1) edge [right] (eq2);
     \path[->]<1-> (eom2) edge [right] (eq2);
     \path[->]<1-> (eom2) edge [right] (eq3);
 \end{tikzpicture}
\end{align*}

但这会产生不完全相同大小的箭头。

我将非常感激任何可以解决此问题的建议。

有没有办法去掉箭头（即只保留线段）？另外，有没有办法确保箭头从文本中心线上的某个点开始？

Answer 1

我看到 Texnician 和我的想法差不多。我实际上去了 tikzmark 并再次尝试将线条居中。添加\strut是关键。

\documentclass{article}
\usepackage{mathtools}
\usepackage{tikzpagenodes}
 \usetikzlibrary{calc}
 \tikzstyle{every picture}+=[remember picture]
 \everymath{\displaystyle}
 %% the above lines go into the preamble
\begin{document}
\begin{equation*}
 \def\arraystretch{2}%
 \begin{array}{ll}
    \tikz[baseline]{\node[anchor=base](eq1){\strut}} \delta\lambda = -\dot{\varphi}\epsilon & +\epsilon\tfrac{\partial}{\partial p}S = \epsilon(\dot{\varphi}-p)\\
    \tikz[baseline]{\node[anchor=base](eq2){\strut}} \delta p = i\epsilon(x\dot{\lambda} + (1-x)2i\dot{\pi}) & -\epsilon\tfrac{\partial}{\partial\lambda}S = \epsilon\dot{\pi}\\
                                            & +\tfrac{i}{2}\epsilon\tfrac{\partial}{\partial \pi}S = -\tfrac{i}{2}\epsilon\dot{\lambda}\\
    \tikz[baseline]{\node[anchor=base](eq3){\strut}} \delta\pi = -\tfrac{i}{2}(-\dot{\varphi}\epsilon) & -\tfrac{i}{2}\epsilon\tfrac{\partial}{\partial p}S = -\tfrac{i}{2}(\dot{\varphi}-p)\epsilon
 \end{array}
\end{equation*}
\begin{tikzpicture}[overlay]
     \path (eq1) -- (eq2) coordinate[midway] (m1);
     \node[right] (eom1) at  (m1 -| current page text area.west) {Text1};
     \path (eq2) -- (eq3) coordinate[midway] (m2);
     \node[right] (eom2) at  (m2 -| current page text area.west) {Text2};
     \draw (eom1.east) -- (eq1);
     \draw (eom1.east) -- (eq2);
     \draw (eom2.east) -- (eq2);
     \draw (eom2.east) -- (eq3);
\end{tikzpicture}