\end{align*} 行中出现 \hbox 溢出错误

Question

\begin{align*}
[C_{\mu\nu}C^{\mu\nu},Q_{\alpha}] 
&= C_{\mu\nu}\eta^{\mu a}\eta^{\nu b}[C_{ab},Q_{\alpha}]+[C_{\mu\nu},Q_{\alpha}]C^{\mu\nu} \\
&= C_{\mu\nu}\eta^{\mu a}\eta^{\nu b}(iP_{b}(\sigma^{ab})_{\alpha}^{\beta}Q_{\beta}P_{b}-iP_{a}(\sigma^{ba})_{\alpha}^{\beta}Q_{\beta}P_{a})+iP_{\nu}(\sigma^{\mu\nu})_{\alpha}^{\beta}Q_{\beta}P_{\nu}C^{\mu\nu}\\
& -iP_{\mu}(\sigma^{\nu\mu})_{\alpha}^{\beta}Q_{\beta}P_{\mu}C^{\mu\nu} \\
&= 0
\end{align*}

Answer 1

通过在术语处进行拆分使第二行更短+ iP_{\nu}(\sigma^{\mu\nu})。

\begin{align*}
[C_{\mu\nu}C^{\mu\nu},Q_{\alpha}] 
&= C_{\mu\nu}\eta^{\mu a}\eta^{\nu b}[C_{ab},Q_{\alpha}]+[C_{\mu\nu},Q_{\alpha}]C^{\mu\nu} \\
&= C_{\mu\nu}\eta^{\mu a}\eta^{\nu b}(iP_{b}(\sigma^{ab})_{\alpha}^{\beta}Q_{\beta}P_{b}-iP_{a}(\sigma^{ba})_{\alpha}^{\beta}Q_{\beta}P_{a}) \\
&\phantom{={}} +iP_{\nu}(\sigma^{\mu\nu})_{\alpha}^{\beta}Q_{\beta}P_{\nu}C^{\mu\nu} - iP_{\mu}(\sigma^{\nu\mu})_{\alpha}^{\beta}Q_{\beta}P_{\mu}C^{\mu\nu} \\
&= 0
\end{align*}

Answer 2

为了分裂第二条方程线，我将使用multlined以下环境mathtools：

\documentclass{article}
\usepackage{mathtools}

\begin{document}
\begin{align*}
[C_{\mu\nu}C^{\mu\nu},Q_{\alpha}]
    & = C_{\mu\nu}\eta^{\mu a}\eta^{\nu b}[C_{ab},Q_{\alpha}]+[C_{\mu\nu},Q_{\alpha}]C^{\mu\nu} \\
    & = \begin{multlined}[t]
        C_{\mu\nu}\eta^{\mu a}\eta^{\nu b}(iP_{b}(\sigma^{ab})_{\alpha}^{\beta}Q_{\beta}P_{b}-iP_{a}(\sigma^{ba})_{\alpha}^{\beta}Q_{\beta}P_{a})\\ +iP_{\nu}(\sigma^{\mu\nu})_{\alpha}^{\beta}Q_{\beta}P_{\nu}C^{\mu\nu} - iP_{\mu}(\sigma^{\nu\mu})_{\alpha}^{\beta}Q_{\beta}P_{\mu}C^{\mu\nu} 
        \end{multlined}    \\
    & = 0
\end{align*}
\end{document}