方程式超出边距。Springer LNCS 格式

Question

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。

Answer 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}