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如何对齐等号

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如何对齐等号

在此处输入图片描述 我想让所有等号对齐。所以我尝试了论坛里的许多解决方案，但都没有效果。我仍然无法让它工作。这是我的代码：

\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel} 
\setlength{\parindent}{4em}
\setlength{\parskip}{1em}
\renewcommand{\baselinestretch}{1.25}
\usepackage{braket}
\usepackage{mathtools}
\everymath{\displaystyle}
\begin{document}
[...]
Where $H\left(\frac{\omega_{0}}{2},\mu\right)$ is the Hamiltonian for isotropic $2-D$ oscillator with frequency $\frac{\omega_{0}}{2}\\
\begin{align}
\left[H\left(\frac{\omega_{0}}{2},\mu\right),L_{z}\right]&=\left[H\left(\frac{\omega_{0}}{2},\mu\right),xp_{y}-yp_{x}\right]\\
\qquad &=\frac{1}{2\mu}\left(\left[p_{x}^2,xp_{y}\right]+\left[p_{y}^2,-yp_{x}\right]\right)+\frac{\mu\omega_{0}^2}{8}\left(\left[x^2,-yp_{x}\right]+\left[y^2,xp_{y}\right]\right)\\
\qquad &=\frac{1}{2\mu}\left[2p_{x}p_{y}(-i\hbar)+2(i\hbar)p_{y}p_{x}\right]+\frac{\mu\omega_{0}^2}{8}\left[2yx(-i\hbar)+2xy(i\hbar)\right]\\
\qquad &=0\textrm{\qquad, }\left([p_{x},p_{y}]=[x,y]=0\right)$\\
\end{align}
[...]
\end{document}

我总是收到这个错误信息：

    ! Package amsmath Error: \begin{align} allowed only in paragraph mode.

    See the amsmath package documentation for explanation.
    Type  H <return>  for immediate help.
     ...                                              

    l.65 \begin{align}

你可以帮帮我吗？

答案1

我假设这all意味着“除了最终向量之外的所有向量”。

代码被更改的行被标记为%here。

\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel} 
\setlength{\parindent}{4em}
\setlength{\parskip}{1em}
\renewcommand{\baselinestretch}{1.25}
\usepackage{braket}
\usepackage{mathtools}
\everymath{\displaystyle}
\begin{document}
[...]
Where $H\left(\frac{\omega_{0}}{2},\mu\right)$ is the Hamiltonian for isotropic $2-D$ oscillator with frequency $\frac{\omega_{0}}{2}$ %here
\begin{align}
\left[H\left(\frac{\omega_{0}}{2},\mu\right),L_{z}\right]&=\left[H\left(\frac{\omega_{0}}{2},\mu\right),xp_{y}-yp_{x}\right]\\
\qquad &=\frac{1}{2\mu}\left(\left[p_{x}^2,xp_{y}\right]+\left[p_{y}^2,-yp_{x}\right]\right)+\frac{\mu\omega_{0}^2}{8}\left(\left[x^2,-yp_{x}\right]+\left[y^2,xp_{y}\right]\right)\\
\qquad &=\frac{1}{2\mu}\left[2p_{x}p_{y}(-i\hbar)+2(i\hbar)p_{y}p_{x}\right]+\frac{\mu\omega_{0}^2}{8}\left[2yx(-i\hbar)+2xy(i\hbar)\right]\\
\qquad &=0\textrm{\qquad, }\left([p_{x},p_{y}]=[x,y]=0\right) %here
\end{align}
[...]
\end{document}

在此处输入图片描述

此外，请注意，\\之前完全不需要类似方程的环境。

答案2

Przemysław Scherwentke 有已回答您的问题，但代码还有一些需要改进：

\documentclass{article}

\usepackage{amsmath}

\everymath{\displaystyle}
\setlength\parindent{4em}

\begin{document}

[\dots]
where $H{\mkern -5mu}\left(\frac{\omega_{0}}{2}, \mu\right){\mkern -5mu}$ is the Hamiltonian for isotropic $\mathrm{2D}$ oscillator with frequency $\frac{\omega_{0}}{2}$;
\begin{align}
  \left[H{\mkern -5mu}\left(\frac{\omega_{0}}{2}, \mu\right){\mkern -5mu}, L_{z}\right]
  &= \left[H{\mkern -5mu}\left(\frac{\omega_{0}}{2}, \mu\right){\mkern -5mu}, xp_{y} - yp_{x}\right]\\
  &= \frac{1}{2\mu}{\mkern -3mu}\left(\left[p_{x}^{2}, xp_{y}\right] + \left[p_{y}^{2}, -yp_{x}\right]\right){\mkern -3mu} + \frac{\mu\omega_{0}^{2}}{8}{\mkern -3mu}\left(\left[x^{2}, -yp_{x}\right] + \left[y^{2}, xp_{y}\right]\right){\mkern -3mu}\\
  &= \frac{1}{2\mu}[2p_{x}p_{y}(-i\hbar) + 2(i\hbar)p_{y}p_{x}] + \frac{\mu\omega_{0}^{2}}{8}[2yx(-i\hbar) + 2xy(i\hbar)]\\
  &= 0,\qquad ([p_{x}, p_{y}] = [x, y] = 0)
\end{align}
[\dots]

\end{document}

输出1

更新

结合 Mico 的建议，得出以下结论：

\documentclass{article}

\usepackage{amsmath}
\usepackage{mleftright}

\everymath{\displaystyle}
\setlength\parindent{4em}

\begin{document}

[\dots]
where $H\mleft(\frac{\omega_{0}}{2}, \mu\mright)$ is the Hamiltonian for isotropic $\mathrm{2D}$ oscillator with frequency $\frac{\omega_{0}}{2}$;
\begin{align}
  \left[H\mleft(\frac{\omega_{0}}{2}, \mu\mright), L_{z}\right]
  &= \mleft[H\mleft(\frac{\omega_{0}}{2}, \mu\mright), xp_{y} - yp_{x}\mright]\\
  &= \frac{1}{2\mu}\mleft(\mleft[p_{x}^{2}, xp_{y}\mright] + \mleft[p_{y}^{2}, -yp_{x}\mright]\right) + \frac{\mu\omega_{0}^{2}}{8}\left(\mleft[x^{2}, -yp_{x}\mright] + \mleft[y^{2}, xp_{y}\mright]\mright)\\
  &= \frac{1}{2\mu}[2p_{x}p_{y}(-i\hbar) + 2(i\hbar)p_{y}p_{x}] + \frac{\mu\omega_{0}^2}{8}[2yx(-i\hbar) + 2xy(i\hbar)]\\
  &= 0,\qquad ([p_{x}, p_{y}] = [x, y] = 0)
\end{align}
[\dots]

\end{document}

输出2

在这里，mleftright添加包是为了删除{\mkern -<x>mu}构造。

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