☰

使用对齐将单个方程编号放入方程的每一列中

tag-icon • tag-icon equations align
使用对齐将单个方程编号放入方程的每一列中

我尝试使用对齐将方程编号放入方程的每一列中。为此，我编写了以下代码-

\begin{align}
\label{equn:rotation}
\begin{split}
W^+_\mu=\frac{W^1_\mu-iW^2_\mu}{\sqrt{2}}\\
W^-_\mu=\frac{W^1_\mu+iW^2_\mu}{\sqrt{2}}
\end{split} & &  \begin{split}
Z^0_\mu=\cos{\theta_w} W^3_\mu-\sin{\theta_w} B_\mu\\
A_\mu=\sin{\theta_w} W^3_\mu+\cos{\theta_w} B_\mu
\end{split}
\end{align}

我得到以下输出-增益输出。

但我想得到这样的输出-预期输出。

但我无法做到这一点。有人能帮我吗？

答案1

我建议使用环境 minipage 的解决方案。希望这对您有所帮助。备注：对于 = 符号的对齐，我认为最好在 split 环境中使用 &=

    \documentclass[11pt]{article}
    \usepackage{amsmath}
        
    \begin{document}
    \begin{minipage}{0.45\linewidth}
        \begin{equation}
            \begin{split}
                W^+_\mu&=\frac{W^1_\mu-iW^2_\mu}{\sqrt{2}}\\
                W^-_\mu&=\frac{W^1_\mu+iW^2_\mu}{\sqrt{2}}
            \end{split}
        \end{equation}%
    \end{minipage}%
    \hfill%
    \begin{minipage}{0.45\linewidth}
        \begin{equation}
            \begin{split}
                Z^0_\mu&=\cos{\theta_w} W^3_\mu-\sin{\theta_w} B_\mu\\
                A_\mu&=\sin{\theta_w} W^3_\mu+\cos{\theta_w} B_\mu
            \end{split}
        \end{equation}%
    \end{minipage}%
    \end{document}

在此处输入图片描述

答案2

我会aligned一起使用mathtools而不是split。

\documentclass[11pt]{article}
\usepackage{mathtools}
\begin{document}
\begin{minipage}{0.45\linewidth}
\begin{equation}
  \begin{aligned}
  W^+_\mu &= \frac{W^1_\mu-iW^2_\mu}{\sqrt{2}}\\
  W^-_\mu &= \frac{W^1_\mu+iW^2_\mu}{\sqrt{2}}
  \end{aligned}
\end{equation}%
\end{minipage}%
\hfill%
\begin{minipage}{0.45\linewidth}
\begin{equation}
  \begin{aligned}
  Z^0_\mu &= \cos{\theta_w} W^3_\mu-\sin{\theta_w} B_\mu\\
  A_\mu   &= \sin{\theta_w} W^3_\mu+\cos{\theta_w} B_\mu
  \end{aligned}
\end{equation}%
\end{minipage}%
\end{document}

答案3

您可以测量这两个部分，为方程式编号添加一些空间，然后使用小页面排版。

可选参数doubleequation用于左缩进。

另一方面，最后一种方式，align看起来更清晰。

\documentclass{article}
\usepackage{amsmath}

\usepackage{lipsum} % for context

\ExplSyntaxOn

\NewDocumentEnvironment{doubleequation}{O{2em}b}
  {
     \anirban_doubleequation:nn { #1 } { #2 }
  }
  {}

\seq_new:N \l__anirban_doubleequation_seq
\dim_new:N \l__anirban_doubleequation_a_dim
\dim_new:N \l__anirban_doubleequation_b_dim
\box_new:N \l__anirban_doubleequation_a_box
\box_new:N \l__anirban_doubleequation_b_box
\box_new:N \l__anirban_doubleequation_eqnum_box

\cs_new_protected:Nn \anirban_doubleequation:nn
  {
     $$ % yes!
     \seq_set_split:Nnn \l__anirban_doubleequation_seq { \nextequation } { #2 }
     \hbox_set:Nn \l__anirban_doubleequation_eqnum_box { \theequation }
     \hbox_set:Nn \l__anirban_doubleequation_a_box
       {
          $\displaystyle\seq_item:Nn \l__anirban_doubleequation_seq { 1 }$
       }
     \dim_set:Nn \l__anirban_doubleequation_a_dim
       { \box_wd:N \l__anirban_doubleequation_a_box + 4em + \box_wd:N \l__anirban_doubleequation_eqnum_box }
     \hbox_set:Nn \l__anirban_doubleequation_b_box
       {
          $\displaystyle\seq_item:Nn \l__anirban_doubleequation_seq { 2 }$
       }
     \dim_set:Nn \l__anirban_doubleequation_b_dim
       { \box_wd:N \l__anirban_doubleequation_b_box + 4em + \box_wd:N \l__anirban_doubleequation_eqnum_box }
     \hspace{#1}
     \begin{minipage}{\l__anirban_doubleequation_a_dim}
       \noindent
       \begin{equation}
       \hspace{0pt}
       \seq_item:Nn \l__anirban_doubleequation_seq { 1 }
       \hspace{10000pt minus 1fil}
       \end{equation}
     \end{minipage}
     \hspace{10000pt minus 1fil}
     \begin{minipage}{\l__anirban_doubleequation_b_dim}
       \noindent
       \begin{equation}
       \hspace{0pt}
       \seq_item:Nn \l__anirban_doubleequation_seq { 2 }
       \hspace{10000pt minus 1fil}
       \end{equation}
     \end{minipage}
     $$
 }

\ExplSyntaxOff

\begin{document}

\eqref{A} and \eqref{B}

\lipsum[1][1-5]
\begin{doubleequation}
  \begin{aligned}
  W^+_\mu&=\frac{W^1_\mu-iW^2_\mu}{\sqrt{2}}\\
  W^-_\mu&=\frac{W^1_\mu+iW^2_\mu}{\sqrt{2}}
  \end{aligned}\label{A}
\nextequation
  \begin{aligned}
  Z^0_\mu&=\cos{\theta_w} W^3_\mu-\sin{\theta_w} B_\mu\\
  A_\mu&=\sin{\theta_w} W^3_\mu+\cos{\theta_w} B_\mu
  \end{aligned}\label{B}
\end{doubleequation}
\lipsum[2][1-5]\setcounter{equation}{123}
\begin{doubleequation}[1em]
  \begin{aligned}
  W^+_\mu&=\frac{W^1_\mu-iW^2_\mu}{\sqrt{2}}\\
  W^-_\mu&=\frac{W^1_\mu+iW^2_\mu}{\sqrt{2}}
  \end{aligned}
\nextequation
  \begin{aligned}
  Z^0_\mu&=\cos{\theta_w} W^3_\mu-\sin{\theta_w} B_\mu\\
  A_\mu&=\sin{\theta_w} W^3_\mu+\cos{\theta_w} B_\mu
  \end{aligned}
\end{doubleequation}
\lipsum[3][1-5]
\begin{align}
 &  W^+_\mu=\frac{W^1_\mu-iW^2_\mu}{\sqrt{2}}
&&  W^-_\mu=\frac{W^1_\mu+iW^2_\mu}{\sqrt{2}}
\\[1ex]
 &  Z^0_\mu=\cos{\theta_w} W^3_\mu-\sin{\theta_w} B_\mu
&&  A_\mu=\sin{\theta_w} W^3_\mu+\cos{\theta_w} B_\mu
\end{align}
\lipsum[4][1-5]

\end{document}

在此处输入图片描述

使用了一些技巧……

相关内容