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使用多线对齐方程

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使用多线对齐方程

你能告诉我一种对齐这些方程的方法吗？

\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{enumerate}
\item \begin{align*}
      \begin{multlined}[t][0.7\textwidth]
      \langle x_{b}, t_{b}|x_{a}, t\rangle -\lim_{\epsilon\to 0} \left( \epsilon \frac{\partial}{\partial t}\langle x_{b}, t_{b}|x_{a}, t\rangle\ \right)\\
       = \lim_{\epsilon\rightarrow 0} \frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\bigg(\langle x_{b}, t_{b}|x_{a}, t\rangle+\xi\frac{\partial}{\partial x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle\\
       +\frac{\xi^{2}}{2}\frac{\partial^{2}}{\partial x^{2}_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle-\frac{iV\epsilon}{\hbar}\langle x_{b}, t_{b}|x_{a}, t\rangle-\frac{iV\xi\epsilon}{\hbar}\frac{\partial}{\partial x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle-\frac{iV\xi^{2}\epsilon}{2\hbar}\frac{\partial^{2}}{\partial x^{2}_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle+\dots\bigg)\\
       = \lim_{\epsilon\rightarrow 0} \bigg[\bigg(\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\langle x_{b}, t_{b}|x_{a}, t\rangle\bigg)\\
       +\bigg(\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} \xi d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\frac{\partial}{\partial x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle\bigg)\\
       +\bigg(\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} \frac{\xi^{2}}{2} d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\frac{\partial^{2}}{\partial x_{a}^{2}}\langle x_{b}, t_{b}|x_{a}, t\rangle\bigg)\\
       -\frac{iV\epsilon}{\hbar}\bigg(\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\langle x_{b}, t_{b}|x_{a}, t\rangle\bigg)\\
       -\frac{iV\epsilon}{\hbar}\bigg(\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} \xi d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\frac{\partial}{\partial x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle\bigg)\\
       -\frac{iV\epsilon}{\hbar}\bigg(\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} \frac{\xi^{2}}{2} d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\frac{\partial^{2}}{\partial x_{a}^{2}}\langle x_{b}, t_{b}|x_{a}, t\rangle\bigg)+\dots\bigg]
      \end{multlined}\\
      \end{align*}
\end{enumerate}
\end{document}

答案1

这是我能提供的；我会删除中间的步骤，这可以从最后一步轻松推断出来。

\documentclass{article}
\usepackage{amsmath,mathtools}

\newcommand{\pihem}{\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}}
\newcommand{\pder}[2][]{\frac{\partial^{#1}}{\partial#2^{#1}}}

\begin{document}

\begin{equation*}
\begin{split}
\mathrlap{
  \langle x_{b}, t_{b}|x_{a}, t\rangle
  -\lim_{\epsilon\to 0} \left( \epsilon \pder{t}\langle x_{b}, t_{b}|x_{a}, t\rangle \right)
}\quad&\\
&= \lim_{\epsilon\rightarrow 0} 
   \pihem \int_{-\infty}^{\infty} d\xi 
   \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)
\\
&\qquad\times
  \biggl(\begin{aligned}[t]
  &\langle x_{b}, t_{b}|x_{a}, t\rangle+\xi\pder{x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle
  \\
  &+\frac{\xi^{2}}{2}\pder[2]{x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle
  -\frac{iV\epsilon}{\hbar}\langle x_{b}, t_{b}|x_{a}, t\rangle
  \\
  &-\frac{iV\xi\epsilon}{\hbar}\pder{x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle
  -\frac{iV\xi^{2}\epsilon}{2\hbar}\pder[2]{x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle
  +\dotsb
  \biggr)\end{aligned}
\\
&= \lim_{\epsilon\rightarrow 0} 
  \biggl[\begin{aligned}[t]
    &\biggl(\pihem \int_{-\infty}^{\infty} d\xi \exp\left(\frac{im\xi}{2\hbar\epsilon}\right)\langle x_{b}, t_{b}|x_{a}, t\rangle\biggr)\\
    &+\biggl(\pihem \int_{-\infty}^{\infty} \xi d\xi \exp\left(\frac{im\xi}{2\hbar\epsilon}\right)\pder{x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle\biggr)\\
    &+\biggl(\pihem \int_{-\infty}^{\infty} \frac{\xi}{2} d\xi \exp\left(\frac{im\xi}{2\hbar\epsilon}\right)\pder[2]{x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle\biggr)\\
    &-\frac{iV\epsilon}{\hbar}\biggl(\pihem \int_{-\infty}^{\infty} d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\langle x_{b}, t_{b}|x_{a}, t\rangle\biggr)\\
    &-\frac{iV\epsilon}{\hbar}\biggl(\pihem \int_{-\infty}^{\infty} \xi d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\pder{x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle\biggr)\\
    &-\frac{iV\epsilon}{\hbar}\biggl(\pihem \int_{-\infty}^{\infty} \frac{\xi^{2}}{2} d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\pder[2]{x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle\biggr)+\dotsb
  \biggr]\end{aligned}
\end{split}
\end{equation*}

\end{document}

请注意\bigg应该是\biggl或\biggr。

在此处输入图片描述

这是没有中间步骤的。

在此处输入图片描述

答案2

在类似的情况下，我将在方程中引入一些新变量，并按以下方式写出它们：

在此处输入图片描述

\documentclass{article}
\usepackage{mathtools}
\usepackage{geometry}

\begin{document}
\begin{enumerate}
\item 
    \begin{align*}
\langle x_{b}, t_{b}|x_{a}, t\rangle 
    &   -\lim_{\epsilon\to 0}\left(\epsilon\frac{\partial}{\partial t}\langle x_{b},t_{b}|x_{a}, t\rangle\ \right)\\
    &   \begin{multlined}[t][0.8\textwidth]
      = \lim_{\epsilon\to 0} 
        \underbrace{\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}}_{=A}
                                \int_{-\infty}^{\infty} d\xi 
        \underbrace{\exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)}_{=B}  \\
                \bigg(\langle x_{b}, t_{b}|x_{a},
                t\rangle+\xi\frac{\partial}{\partial x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle 
                + \frac{\xi^{2}}{2}\frac{\partial^{2}}{\partial x^{2}_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle 
                -\frac{iV\epsilon}{\hbar}\langle x_{b}, t_{b}|x_{a}, t\rangle \\
                -\frac{iV\xi\epsilon}{\hbar}\frac{\partial}{\partial x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle
                -\frac{iV\xi^{2}\epsilon}{2\hbar}\frac{\partial^{2}}{\partial x^{2}_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle+\dots\bigg)
      \end{multlined}       \\
    &  \begin{multlined}[t][0.8\textwidth]
       = \lim_{\epsilon\to 0} \bigg[
        A\int_{-\infty}^{\infty} d\xi  B \langle x_{b}, t_{b}|x_{a}, t\rangle 
        + A \int_{-\infty}^{\infty} \xi d\xi B 
                \frac{\partial}{\partial x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle     \\
       + A \int_{-\infty}^{\infty} \frac{\xi^{2}}{2} d\xi B 
                \frac{\partial^{2}}{\partial x_{a}^{2}}\langle x_{b}, t_{b}|x_{a}, t\rangle 
       - \frac{iV\epsilon}{\hbar} A\int_{-\infty}^{\infty} d\xi B
                \langle x_{b}, t_{b}|x_{a}, t\rangle    \\
       - \frac{iV\epsilon}{\hbar} A \int_{-\infty}^{\infty} \xi d\xi B
                \frac{\partial}{\partial x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle 
       -\frac{iV\epsilon}{\hbar} A \int_{-\infty}^{\infty} \frac{\xi^{2}}{2} d\xi B \frac{\partial^{2}}{\partial x_{a}^{2}}\langle x_{b}, t_{b}|x_{a}, t\rangle \\ 
       + \dots \bigg]
      \end{multlined}
      \end{align*}
\end{enumerate}
\end{document}

答案3

仍然不太可读，但是 在此处输入图片描述

\documentclass{article}
\usepackage{mathtools}
\begin{document}


\begin{equation}
      \begin{aligned}
      \shoveleft \langle x_{b}, t_{b}|x_{a}, t\rangle -\lim_{\epsilon\to 0} \left( \epsilon \frac{\partial}{\partial t}\langle x_{b}, t_{b}|x_{a}, t\rangle\ \right)
\kern-15em\\
&= \lim_{\epsilon\rightarrow 0} \frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\bigg(\langle x_{b}, t_{b}|x_{a}, t\rangle\\
&\qquad+\xi\frac{\partial}{\partial x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle
    +\frac{\xi^{2}}{2}\frac{\partial^{2}}{\partial x^{2}_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle\\
&\qquad-\frac{iV\epsilon}{\hbar}\langle x_{b}, t_{b}|x_{a}, t\rangle-\frac{iV\xi\epsilon}{\hbar}\frac{\partial}{\partial x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle\\
&\qquad-\frac{iV\xi^{2}\epsilon}{2\hbar}\frac{\partial^{2}}{\partial x^{2}_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle+\dots\bigg)\\
       &= \lim_{\epsilon\rightarrow 0} \bigg[\bigg(\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\langle x_{b}, t_{b}|x_{a}, t\rangle\bigg)\\
&       +\bigg(\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} \xi d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\frac{\partial}{\partial x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle\bigg)\\
&       +\bigg(\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} \frac{\xi^{2}}{2} d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\frac{\partial^{2}}{\partial x_{a}^{2}}\langle x_{b}, t_{b}|x_{a}, t\rangle\bigg)\\
&       -\frac{iV\epsilon}{\hbar}\bigg(\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\langle x_{b}, t_{b}|x_{a}, t\rangle\bigg)\\
&       -\frac{iV\epsilon}{\hbar}\bigg(\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} \xi d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\frac{\partial}{\partial x_{a}}\langle x_{b}, t_{b}|x_{a}, t\rangle\bigg)\\
&       -\frac{iV\epsilon}{\hbar}\bigg(\frac{1}{\sqrt{2\pi i\hbar\epsilon/m}}\int_{-\infty}^{\infty} \frac{\xi^{2}}{2} d\xi \exp\left(\frac{im\xi^{2}}{2\hbar\epsilon}\right)\frac{\partial^{2}}{\partial x_{a}^{2}}\langle x_{b}, t_{b}|x_{a}, t\rangle\bigg)\\
&\qquad+\dots\bigg]
      \end{aligned}
      \end{equation}

\end{document}

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