![方程式超出边距。Springer LNCS 格式](https://linux22.com/image/438812/%E6%96%B9%E7%A8%8B%E5%BC%8F%E8%B6%85%E5%87%BA%E8%BE%B9%E8%B7%9D%E3%80%82Springer%20LNCS%20%E6%A0%BC%E5%BC%8F%20.png)
我使用的公式超出了 Springer LNCS 格式的边距。MWE 如下所示。
\documentclass[runningheads]{llncs}
\usepackage{amsmath,amssymb}
\usepackage{mathtools, nccmath}
\begin{document}
\begin{align}
L_{B_{1} A_{1}} L_{B_{0} A_{0}}\left(\left|I_{1}\right\rangle\right) & =L_{B_{1} A_{1}} L_{B_{0} A_{0}}\left(\frac{1}{2^{m}} \sum_{b=0}^{2^{m}-1} \sum_{a=0}^{2^{m}-1}\left|P_{b a}\right\rangle\left|b^{\prime}\right\rangle\left|a^{\prime}\right\rangle\right)\\
& =\frac{1}{2^{m}}\left(\underset{b a \neq B_{0} A_{0}, B_{1} A_{1}}{\displaystyle\sum_{b=0}^{2^{m}-1} \sum_{a=0}^{2^{m}-1}}\left|P_{ba}\right\rangle\left|b^{\prime}\right\rangle\left|a^{\prime}\right\rangle+U_{B A}\left(\left|P_{B_{0} A_{0}}\right|\right)\left|B_{0} A_{0}\right\rangle+U_{B A}\left(\left|P_{B_{1} A_{1}}\right\rangle\right)\left|B_{1} A_{1}\right\rangle\right) \\
& =\frac{1}{2^{m}}
\left(\begin{array}{rl}
\underset{b a \neq B_{0} A_{0}, B_{1} A_{1}}{\displaystyle\sum_{b=0}^{2^{m}-1} \sum_{a=0}^{2^{m}-1}}\left|P_{b a}\right\rangle\left|b^{\prime}\right\rangle\left|a^{\prime}\right\rangle &+\left|p_{B_{0} A_{0}}^{2} p_{B_{0} A_{0}}^{3^{\prime}} \cdots p_{B_{0} A_{0}}^{0^{\prime}} p_{B_{0} A_{0}}^{1^{\prime}}\right\rangle\left|B_{0} A_{0}\right\rangle \\
&+\left|p_{B_{1} A_{1}}^{2^{\prime}} p_{B_{1} A_{1}}^{3^{\prime}} \cdots p_{B_{1} A_{1}}^{0^{\prime}} p_{B_{1} A_{1}}^{1^{\prime}}\right\rangle\left|B_{1} A_{1}\right\rangle
\end{array}\right)
\end{align}
\end{document}
假设假想的黄线是边界。你可以看到方程式越过了这个边界。每当我尝试时,它都会显示错误。请有人帮我改正它。
答案1
我建议使用fleqn
环境,因为您需要加载nccmath
和aligned
环境。请注意,您不必加载 amsmath
,因为amsmath
和都nccmath
加载它。
\documentclass[runningheads]{llncs}
\usepackage{amssymb}
\usepackage{nccmath, mathtools}
\begin{document}
\begin{fleqn}
\begin{align}
& L_{B_{1} A_{1}} L_{B_{0} A_{0}}\left(\left|I_{1}\right\rangle\right) =L_{B_{1} A_{1}} L_{B_{0} A_{0}}\left(\frac{1}{2^{m}} \sum_{b=0}^{2^{m}-1} \sum_{a=0}^{2^{m}-1}\left|P_{b a}\right\rangle\left|b^{\prime}\right\rangle\left|a^{\prime}\right\rangle\right)\\
& =\frac{1}{2^{m}}\Biggl(\underset{b a \neq B_{0} A_{0}, B_{1} A_{1}}{\displaystyle\sum_{b=0}^{2^{m}-1} \sum_{a=0}^{2^{m}-1}}\left|P_{ba}\right\rangle\left|b^{\prime}\right\rangle\left|a^{\prime}\right\rangle \begin{aligned}[t] & +U_{B A}\left(\left|P_{B_{0} A_{0}}\right|\right)\left|B_{0} A_{0}\right\rangle\\
& +U_{B A}\left(\left|P_{B_{1} A_{1}}\right\rangle\right)\left|B_{1} A_{1}\right\rangle\Biggr)
\end{aligned}\\
& =\frac{1}{2^{m}}
\Biggl( \underset{b a \neq B_{0} A_{0}, B_{1} A_{1}}{\sum_{b=0}^{2^{m}-1} \sum_{a=0}^{2^{m}-1}}\hspace*{-0.5em}\left|P_{b a}\right\rangle\left|b^{\prime}\right\rangle\left|a^{\prime}\right\rangle
\begin{aligned}[t] &+\left|p_{B_{0} A_{0}}^{2} p_{B_{0} A_{0}}^{3^{\prime}} \cdots p_{B_{0} A_{0}}^{0^{\prime}} p_{B_{0} A_{0}}^{1^{\prime}}\right\rangle\left|B_{0} A_{0}\right\rangle \\
&+\left|p_{B_{1} A_{1}}^{2^{\prime}} p_{B_{1} A_{1}}^{3^{\prime}} \cdots p_{B_{1} A_{1}}^{0^{\prime}} p_{B_{1} A_{1}}^{1^{\prime}}\right\rangle\left|B_{1} A_{1}\right\rangle\Biggr)
\end{aligned}
\end{align}
\end{fleqn}
\end{document}