\documentclass{article}
\usepackage{amsmath}
\usepackage{breqn}
\begin{document}
\begin{equation}
F^{(2)}_0(\chi, \Theta, \phi) &= \frac{1}{4}h^{(2)}(j_i,j_f)(3 \sin^2 \Theta \cos 2\chi-[3 \cos^2 \Theta -1])
\vspace{4 mm}
{\em F}^{(2)}_{1+}(\chi, \Theta, \phi) &= \frac{\sqrt{3}}{4}h^{(2)}(j_i,j_f)c_2(j_i)(2 \sin \Theta \cos \phi \sin 2 \chi
+ 2 \sin \Theta \cos \Theta \sin \phi \cos 2 \chi - \sin 2 \Theta \cos \phi)
\vspace{4 mm}
{\em F}^{(2)}_{2+}(\chi, \Theta, \phi) &= \frac{\sqrt{3}}{4}h^{(2)}(j_i,j_f)c_2(j_i)([1 + \cos^2 \Theta] \cos 2\phi \cos 2\chi - 2 \cos \Theta \sin 2\phi \sin 2\chi {-}\sin^2\Theta \cos 2 \phi)\\
\vspace{2 mm}
\end{equation}
\end{document}
答案1
该文件产生
! Missing $ inserted.
<inserted text>
$
l.11
?
! Display math should end with $$.
<to be read again>
\par
l.11
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! Missing $ inserted.
<inserted text>
$
l.12 F^{(2)}_0(\chi
, \Theta, \phi) &= \frac{1}{4}h^{(2)}(j_i,j_f)(3 \sin^2 \...
?
! Misplaced alignment tab character &.
l.12 F^{(2)}_0(\chi, \Theta, \phi) &
= \frac{1}{4}h^{(2)}(j_i,j_f)(3 \sin^2 \...
出现错误后,tex 仅能恢复足够的内容来尝试检查文档的后续部分,排版输出通常根本无法使用。
因此您应该询问错误,而不是询问 pdf 结果。
您不能有空行,equation
并且&
对齐的语法align
也不是equation
这样的:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align}
F^{(2)}_0(\chi, \Theta, \phi) &= \frac{1}{4}h^{(2)}(j_i,j_f)(3 \sin^2 \Theta \cos 2\chi-[3 \cos^2 \Theta -1])\\
\begin{split}F^{(2)}_{1+}(\chi, \Theta, \phi) &= \frac{\sqrt{3}}{4}h^{(2)}(j_i,j_f)c_2(j_i)(2 \sin \Theta \cos \phi \sin 2 \chi
+\\
&\qquad 2 \sin \Theta \cos \Theta \sin \phi \cos 2 \chi - \sin 2 \Theta \cos \phi)
\end{split}\\
\begin{split}F^{(2)}_{2+}(\chi, \Theta, \phi) &= \frac{\sqrt{3}}{4}h^{(2)}(j_i,j_f)c_2(j_i)([1 + \cos^2 \Theta] \cos 2\phi \cos 2\chi -
\\
&\qquad 2 \cos \Theta \sin 2\phi \sin 2\chi {-}\sin^2\Theta \cos 2 \phi)
\end{split}
\end{align}
\end{document}
答案2
我建议您使用单个align
环境、4 个显式\\
(换行)指令和两个\notag
指令。如果您缺少(垂直)空间,我建议您将这三个\frac
指令替换为\tfrac
。请注意,环境中不能有空行align
。
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align}
F^{(2)}_0(\chi, \Theta, \phi)
&= \tfrac{1}{4}h^{(2)}(j_i,j_f)\bigl(3 \sin^2\Theta \cos 2\chi
-[3 \cos^2\Theta -1]\bigr)\\
F^{(2)}_{1+}(\chi, \Theta, \phi)
&= \tfrac{1}{4}\sqrt{3}h^{(2)}(j_i,j_f)c_2(j_i)\bigl(
2\sin \Theta \cos \phi \sin 2\chi \notag\\
&\qquad+ 2\sin \Theta \cos \Theta \sin \phi \cos 2\chi
- \sin 2\Theta \cos \phi\bigr)\\
F^{(2)}_{2+}(\chi, \Theta, \phi) &= \tfrac{1}{4}\sqrt{3}h^{(2)}
(j_i,j_f)c_2(j_i)\bigl([1 + \cos^2\Theta] \cos 2\phi \cos 2\chi \notag\\
&\qquad- 2\cos \Theta \sin 2\phi \sin 2\chi -\sin^2\Theta \cos 2\phi\bigr)
\end{align}
\end{document}