我有以下代码:
\begin{equation}
\begin{aligned}
R^{1}_{\mu\nu} = \int & \frac{d^4p}{(2\pi)^4} \tr \left[ \frac{1}{\slashed{p} - \slashed{k}_2-m} \gamma_{5} \gamma_{\nu} \frac{1}{\slashed{p} - \slashed{q}-m} \gamma_{\mu} \\
& - \frac{1}{\slashed{p} -m} \gamma_{5} \gamma_{\nu} \frac{1}{\slashed{p} - \slashed{k}_1-m} \gamma_{\mu} \right]
\end{aligned}
\end{equation}
以下是我得到的结果
如您所见,我使用了一个长方程,该方程通过 \begin{aligned} 一分为二,第一行中有一个 \left[,它显示了,但第二行中有一个 \right],它没有显示在最终方程上。您知道可能出了什么问题吗?
答案1
为了能够在带换行符的等式中使用\left[
和,您必须使用\right]
包裹breqn
:
\documentclass{article}
\usepackage{amsmath}
\usepackage{slashed}
\usepackage{breqn}
\newcommand*{\tr}{\ensuremath{\mathrm{tr}}}
\begin{document}
\begin{dmath}
R^{1}_{\mu\nu} = \int \frac{d^4p}{(2\pi)^4} \tr \left[ \frac{1}{\slashed{p} - \slashed{k}_2-m} \gamma_{5} \gamma_{\nu} \frac{1}{\slashed{p} - \slashed{q}-m} \gamma_{\mu} \\
- \frac{1}{\slashed{p} -m} \gamma_{5} \gamma_{\nu} \frac{1}{\slashed{p} - \slashed{k}_1-m} \gamma_{\mu} \right]
\end{dmath}
\end{document}
另一种方法是使用\right.
和\left.
,但存在大小不同的风险:
\documentclass{article}
\usepackage{amsmath}
\usepackage{slashed}
\newcommand*{\tr}{\ensuremath{\mathrm{tr}}}
\begin{document}
\begin{equation}
\begin{aligned}
R^{1}_{\mu\nu} = \int & \frac{d^4p}{(2\pi)^4} \tr \left[
\frac{1}{\slashed{p} - \slashed{k}_2-m}
\gamma_{5} \gamma_{\nu} \frac{1}{\slashed{p} -
\slashed{q}-m} \gamma_{\mu} \right. \\
& \left. - \frac{1}{\slashed{p} -m} \gamma_{5} \gamma_{\nu} \frac{1}{\slashed{p} - \slashed{k}_1-m} \gamma_{\mu} \right]
\end{aligned}
\end{equation}
\end{document}
或者明确使用尺寸:
\documentclass{article}
\usepackage{amsmath}
\usepackage{slashed}
\newcommand*{\tr}{\ensuremath{\mathrm{tr}}}
\begin{document}
\begin{equation}
\begin{aligned}
R^{1}_{\mu\nu} = \int & \frac{d^4p}{(2\pi)^4} \tr \biggl[
\frac{1}{\slashed{p} - \slashed{k}_2-m}
\gamma_{5} \gamma_{\nu} \frac{1}{\slashed{p} -
\slashed{q}-m} \gamma_{\mu} \\
& - \frac{1}{\slashed{p} -m} \gamma_{5} \gamma_{\nu} \frac{1}{\slashed{p} - \slashed{k}_1-m} \gamma_{\mu} \biggr]
\end{aligned}
\end{equation}
\end{document}