如何使拆分方程的第二行出现正确的期望符号进入对齐环境

Question

下面是 Latex 代码片段的渲染：我无法在拆分方程的第二行（在 eq(3) 和 eq(4) 上）显示正确的期望符号（ \right> ）。

\noindent To calculate the variance we assume ${a_{\ell m, p h}^2}$ is constant in order to not have a random denominator:
\begin{align} 
\operatorname{Var}\left(\hat{o}_{\ell}\right) &=\dfrac{1}{(2 \ell+1)^2C_{\ell, p h}^2 f_{\text{sky}}} \left<\left(\sum_{m=-\ell}^{\ell} (|b_{\ell m}|^2 - B_\ell) +  2Re(a_{\ell m}^*b_{\ell m}) \right)^2 \right> \nonumber \\
&=\alpha \sum_{m,m' =-\ell}^{\ell} 
\left<\left(b_{\ell m}b_{\ell m}^* - B_\ell + a_{\ell m}^*b_{\ell m}+a_{\ell m}b_{\ell m}^* \right)\nonumber \\
&\quad\quad\left(b_{\ell m'}b_{\ell m'}^* - B_\ell +  a_{\ell m'}^*b_{\ell m'}+a_{\ell m'}b_{\ell m'}^*\right)\right>
\end{align}
\text{with $=\alpha= \dfrac{1}{(2 \ell+1)^2C_{\ell, p h}^2 f_{\text{sky}}}$}
The terms like $B_\ell \left< a\dots b \dots\right>$ or $\left<bb\dots a\dots b \dots\right>$ where $a_{\ell m}$  appears only once are 0 since a is centered and independent of the rest. After simplification and rearranging of the terms we obtain : 
\begin{align}
\operatorname{Var}\left(\hat{o}_{\ell}\right)&=\alpha \sum_{m,m' =-\ell}^{\ell} 
\left<b_{\ell m}b_{\ell m}^*b_{\ell m'}b_{\ell m'}^*\right> - B_\ell^2 +\left<a_{\ell m}^*b_{\ell m}a_{\ell m'}^*b_{\ell m'}\nonumber \\
&+a_{\ell m}b_{\ell m}^*a_{\ell m'}^*b_{\ell m'}+a_{\ell m}^*b_{\ell m}a_{\ell m'}b_{\ell m'}^*+a_{\ell m}b_{\ell m}^*a_{\ell m'}b_{\ell m'}^*\right> \nonumber \\
\end{align}

这里是渲染的捕获：

如何使这些缺失的\right>期望符号出现？

Answer 1

您的代码很好地提醒了过度使用\left和的弊端\right。我会使用明确的大小指令，尤其是、\bigl和。事实上，下面显示的代码不包含\bigr\biggl\biggr任何\left和的实例\right。

另外，我会使用合理放置的^{}粒子（“空指数”）来确保所有下标项都放置在统一的深度。哦，请不要将其用于\dfrac内联数学表达式。

\documentclass{article} % or some other suitable document class
\usepackage[T1]{fontenc}
\usepackage{mathtools} % load 'amsmath' automatically
\DeclareMathOperator{\Var}{Var} % variance operator
\usepackage{microtype}

\begin{document}
\noindent 
To calculate the variance we assume $a^2_{\ell m, p h}$ 
is constant in order to not have a random denominator:
\begin{align} 
\Var(\hat{o}_{\ell}) 
&=\frac{1}{(2 \ell+1)^2C_{\ell, p h}^2 f^{}_{\mathrm{sky}}} 
  \biggl<\biggl(\,\sum_{m=-\ell}^{\ell} 
    (|b^{}_{\ell m}|^2 - B^{}_{\ell}) 
    +2\Re(a^*_{\ell m}b^{}_{\ell m}) \biggr)^{\!2}\, 
  \biggr> \nonumber \\
&=\alpha \sum_{\mathclap{m,m' =-\ell}}^{\ell} 
\bigl\langle (b^{}_{\ell m}b^*_{\ell m} - B^{}_{\ell} 
    + a^*_{\ell m}  b^{}_{\ell m}
    + a^{}_{\ell m} b^*_{\ell m} ) \nonumber \\
&\qquad\qquad\times(b^{}_{\ell m'}b^*_{\ell m'} - B^{}_{\ell} 
  +a^*_{\ell m'}b^{}_{\ell m'} 
  +a^{}_{\ell m'}b^*_{\ell m'}) \bigr\rangle
\end{align}
with $\alpha= 1/\bigl[(2 \ell+1)^2C_{\ell, p h}^2 f^{}_{\mathrm{sky}}\bigr]$.
The terms like $B^{}_{\ell} \langle a\dots b \dots\rangle$ 
or $\langle bb\dots a\dots b \dots\rangle$ 
in which $a^{}_{\ell m}$ appears only once are $0$ 
since $a$ is centered and independent of the rest. After 
simplification and rearranging of the terms we obtain: 
\begin{align}
\Var(\hat{o}_{\ell})
&=\alpha \sum_{\mathclap{m,m' =-\ell}}^{\ell} 
\langle b^{}_{\ell m}b^*_{\ell m} b^{}_{\ell m'}b^*_{\ell m'}\rangle 
  - B^2_{\ell} +\bigl\langle 
        a^*_{\ell m} b^{}_{\ell m}a^*_{\ell m'} b^{}_{\ell m'} \nonumber \\
&\qquad+a^{}_{\ell m}b^*_{\ell m} a^*_{\ell m'} b^{}_{\ell m'} 
       +a^*_{\ell m} b^{}_{\ell m}a^{}_{\ell m'}b^*_{\ell m'} 
       +a^{}_{\ell m}b^*_{\ell m} a^{}_{\ell m'}b^*_{\ell m'}\bigr\rangle  
\end{align}
\end{document}