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聚集-拆分方程

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聚集-拆分方程

请问方程式没有被拆分两次的问题是什么？

   \documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[english]{babel}
\usepackage{amsmath}
\usepackage{physics}
\usepackage{mathtools}
\usepackage{calc} 
\usepackage{float}
\usepackage{fancyhdr}
\usepackage{physics}
\usepackage{lmodern}
\usepackage{mathtools}
\usepackage[makeroom]{cancel}
\usepackage{mdframed}
\begin{document}
\begin{equation}\label{sphi}
\begin{gathered}
\expval**{\frac{e'^2}{|\vec{r_1}-\vec{r_2}|}}{R_{1s}^{(1)} R_{1s}^{(2)} Y_{00}^{(1)} Y_{00}^{(2)}}= \\ =  4 \pi  e'^2 \expval**{\frac{1}{r_2} \sum\limits_{l=0}^{\infty} \sum\limits_{m=-l}^l \left(\frac{r_1}{r_2} \right)^l \cdot \frac{Y^*_{lm}(\vec{n}_1) Y_{lm}(\vec{n}_2)}{2l+1}}{R_{1s}^{(1)} R_{1s}^{(2)} Y_{00}^{(1)} Y_{00}^{(2)}}
\end{gathered} = \\ = 4 \pi  e'^2 \sum\limits_{l=0}^{\infty} \frac{1}{2l+1} \sum\limits_{m=-l}^l \int Y_{00}^*(\vec{n}_1 Y_{lm}^*(\vec{n}_1 Y_{00}(\vec{n}_1 \mathrm{d} \Omega_1 \cdot \int Y_{00}^*(\vec{n}_2 Y_{lm}^*(\vec{n}_2 Y_{00}(\vec{n}_2 \mathrm{d} \Omega_2 \cdot \int_{r_1=0}^{\infty} \int_{r_2=0}^{\infty} r_1^2 r_2^2 R_{1s}^{2(1)} R_{1s}^{2(2)} \cdot \frac{r_1^l}{r_2^{l+1}} \mathrm{d}r_1 \mathrm{d}r_2.
\end{gathered)
\end{equation}
\end{document}

在此处输入图片描述

答案1

这是一种可能的布局：

  \documentclass[12pt, a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage[english]{babel}
\usepackage{physics}
\usepackage{mathtools, nccmath}
\usepackage{calc}
\usepackage{float}
\usepackage{fancyhdr}
\usepackage[makeroom]{cancel}
\usepackage{mdframed}
\usepackage[showframe]{geometry}

\begin{document}

\mbox{}
\begin{fleqn}
\begin{equation}\label{sphi}
\begin{aligned}[b]
 & \expval**{\frac{e'^2}{|\vec{r_1}-\vec{r_2}|}}{R_{1s}^{(1)} R_{1s}^{(2)} Y_{00}^{(1)} Y_{00}^{(2)}}= \raisetag{4.75ex}\\
 & = 4 \pi e'^2 \expval**{\frac{1}{r_2} \sum\limits_{l=0}^{\infty} \sum\limits_{m=-l}^l \left(\frac{r_1}{r_2} \right)^l \cdot \frac{Y^*_{lm}(\vec{n}_1) Y_{lm}(\vec{n}_2)}{2l+1}}{R_{1s}^{(1)} R_{1s}^{(2)} Y_{00}^{(1)} Y_{00}^{(2)}} \\
 & = \begin{aligned}[t]4 \pi e'^2 \sum\limits_{l=0}^{\infty} \frac{1}{2l+1} \sum\limits_{m=-l}^l \int Y_{00}^*(\vec{n}_1 Y_{lm}^*(\vec{n}_1 Y_{00}(\vec{n}_1 \mathrm{d} \Omega_1 \cdot\int Y_{00}^*(\vec{n}_2 Y_{lm}^*(\vec{n}_2 Y_{00}(\vec{n}_2 \mathrm{d} \Omega_2 \\ \cdot\int_{r_1=0}^{\infty}\int_{\mathrlap{r_2=0}}^{\infty} r_1^2 r_2^2 R_{1s}^{2(1)} R_{1s}^{2(2)} \cdot \frac{r_1^l}{r_2^{l+1}} \mathrm{d}r_1 \mathrm{d}r_2.%
 \end{aligned}
\end{aligned}
\end{equation}
\end{fleqn}

\end{document}

在此处输入图片描述

答案2

您正在使用的equation是单行方程，因此\\之间的=不执行任何操作，请使用多行显示环境，例如gather

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