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嵌套子方程的间距/换行建议

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嵌套子方程的间距/换行建议

我的工作示例是：

\documentclass[a4paper, 12pt]{scrreprt}

\usepackage{mathrsfs}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{braket}
\usepackage{mleftright}

\begin{document}

Due to:
\begin{equation}
\label{FourierRelationPositionMomentumKets}
\braket{x' \vert p'} = \frac{1}{\sqrt{2 \pi \hbar}} \exp \mleft( \frac{i p' x'}{\hbar} \mright)
\end{equation}
they are related by Fourier transformation:
\begin{subequations}
\begin{subequations}
    \label{Position-MomentumSpaceWaveFunctionsDiscr}
    \begin{align}
        \psi \mleft( x \mright) = \braket{x \vert \psi} = \braket{x \vert \hat{\mathbb{I}} \vert \psi} = \braket{x \vert \mleft( \sum_{p' \in \mathscr{P}} \ket{p'} \bra{p'} \mright) \vert \psi} = \sum_{p' \in \mathscr{P}} \braket { x \vert p' } \braket{ p' \vert \psi } = \sum_{p' \in \mathscr{P}} \braket { x \vert p' } \phi \mleft( p' \mright), \label{PositionSpaceWaveFunctionsDiscr} \\
        \phi \mleft( p \mright) = \braket{p \vert \psi} = \braket{p \vert \hat{\mathbb{I}} \vert \psi} = \braket{p \vert \mleft( \sum_{x' \in \mathscr{X}} \ket{x'} \bra{x'} \mright) \vert \psi} = \sum_{x' \in \mathscr{X}} \braket { p \vert x' } \braket{ x' \vert \psi } = \sum_{x' \in \mathscr{X}} \braket { p \vert x' } \psi \mleft( x' \mright) \label{MomentumSpaceWaveFunctionsDiscr}
    \end{align}
\end{subequations}
or:
\begin{subequations}
    \label{Position-MomentumSpaceWaveFunctionsCont}
    \begin{align}
        \psi \mleft( x \mright) = \braket{x \vert \psi} = \braket{x \vert \hat{\mathbb{I}} \vert \psi} = \braket{x \vert \mleft( \int\limits_{\mathscr{P}} dp' \, \ket{p'} \bra{p'} \mright) \vert \psi} = \int\limits_{\mathscr{P}} dp' \, \braket { x \vert p' } \braket{ p' \vert \psi } = \int\limits_{\mathscr{P}} dp' \, \braket { x \vert p' } \phi \mleft( p' \mright), \label{PositionSpaceWaveFunctionsCont} \\
        \phi \mleft( p \mright) = \braket{p \vert \psi} = \braket{p \vert \hat{\mathbb{I}} \vert \psi} = \braket{p \vert \mleft( \int\limits_{\mathscr{X}} dx' \, \ket{x'} \bra{x'} \mright) \vert \psi} = \int\limits_{\mathscr{X}} dx' \, \braket { p \vert x' } \braket{ x' \vert \psi } = \int\limits_{\mathscr{X}} dx' \, \braket { p \vert x' } \psi \mleft( x' \mright), \label{MomentumSpaceWaveFunctionsCont}
    \end{align}
\end{subequations}
\end{subequations}

\end{document}

它产生以下输出： 在此处输入图片描述 *我截屏后把傅里叶变换中的 F 大写。

我的问题实际上是方程式 0.2。我希望它们以它们原来的方式嵌套，但它们显然不适合，但我认为将它们分开会使它变得非常庞大。有什么建议可以让我真正节省整个文档的这一部分的空间吗？

答案1

我会保持对齐并将内容分成两行，这样也许最重要的步骤就直接位于彼此下方。我还建议使用直立的差异，并在标点符号前留出空格。太多不必要的错误发生了，因为有人误认为 是$x_i,$。$x_{i'}$我不太喜欢嵌套，subequations但你似乎喜欢它。

\documentclass[a4paper, 12pt,fleqn]{scrreprt}

\usepackage{mathrsfs}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{braket}
\usepackage{mleftright}
\newcommand{\diff}{\mathop{}\!\mathrm{d}}
\begin{document}

Due to:
\begin{equation}
\label{FourierRelationPositionMomentumKets}
\braket{x' \vert p'} = \frac{1}{\sqrt{2 \pi \hbar}} 
\exp \mleft( \frac{\mathrm{i}\, p' x'}{\hbar} \mright)
\end{equation}
they are related by Fourier transformation:
\begin{subequations}
\begin{subequations}
\label{Position-MomentumSpaceWaveFunctionsDiscr}
\begin{align}
    \psi \mleft( x \mright) = \braket{x \vert \psi} = \braket{x \vert \hat{\mathbb{I}} \vert \psi} 
    &= \braket{x \vert \mleft( \sum_{p' \in \mathscr{P}} \ket{p'} \bra{p'} \mright) \vert \psi} 
    \notag\\
    &= \sum_{p' \in \mathscr{P}} \braket { x \vert p' } \braket{ p' \vert \psi }
    = \sum_{p' \in \mathscr{P}} \braket { x \vert p' } \phi \mleft( p'
    \mright)\;, \label{PositionSpaceWaveFunctionsDiscr} \\
    \phi \mleft( p \mright) = \braket{p \vert \psi} = \braket{p \vert \hat{\mathbb{I}} \vert \psi} 
    &= \braket{p \vert \mleft( \sum_{x' \in \mathscr{X}} \ket{x'} \bra{x'} \mright) \vert \psi} 
    \notag\\
    &= \sum_{x' \in \mathscr{X}} \braket { p \vert x' } \braket{ x' \vert \psi } = \sum_{x' \in \mathscr{X}} \braket { p \vert x' } \psi \mleft( x' \mright) \label{MomentumSpaceWaveFunctionsDiscr}
\end{align}
\end{subequations}
or
\begin{subequations}
\label{Position-MomentumSpaceWaveFunctionsCont}
\begin{align}
    \psi \mleft( x \mright) = \braket{x \vert \psi} = \braket{x \vert
    \hat{\mathbb{I}} \vert \psi} &= \braket{x \vert \mleft(
    \int\limits_{\mathscr{P}} \!\diff p' \, \ket{p'} \bra{p'} \mright) \vert \psi} 
    \notag\\
    &= \int\limits_{\mathscr{P}}\! \diff p' \, \braket { x \vert p' } \braket{ p' \vert
    \psi } = \int\limits_{\mathscr{P}}\! \diff p' \, \braket { x \vert p' } \phi
    \mleft( p' \mright)\;, \label{PositionSpaceWaveFunctionsCont} \\
    \phi \mleft( p \mright) = \braket{p \vert \psi} = \braket{p \vert \hat{\mathbb{I}} \vert \psi} 
    &= \braket{p \vert \mleft( \int\limits_{\mathscr{X}}\! \diff x' \, \ket{x'} \bra{x'} \mright) \vert \psi} 
    \notag\\
    &= \int\limits_{\mathscr{X}}\! \diff x' \, \braket { p \vert x' } \braket{ x' \vert
    \psi } = \int\limits_{\mathscr{X}}\! \diff x' \, \braket { p \vert x' } \psi
    \mleft( x' \mright)\;, \label{MomentumSpaceWaveFunctionsCont}
\end{align}
\end{subequations}
\end{subequations}

\end{document}

在此处输入图片描述

答案2

像这样？

在此处输入图片描述

使用split：

\documentclass[a4paper, 12pt]{scrreprt}

\usepackage{mathrsfs}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{braket}
\usepackage{mleftright}

\begin{document}

Due to:
\begin{equation}
\label{FourierRelationPositionMomentumKets}
\braket{x' \vert p'} = \frac{1}{\sqrt{2 \pi \hbar}} \exp \mleft( \frac{i p' x'}{\hbar} \mright)
\end{equation}
they are related by Fourier transformation:
\begin{subequations}
\begin{subequations}
    \label{Position-MomentumSpaceWaveFunctionsDiscr}
    \begin{align}
\begin{split}\label{PositionSpaceWaveFunctionsDiscr} 
\psi \mleft( x \mright) 
    & = \braket{x \vert \psi} 
      = \braket{x \vert \hat{\mathbb{I}} \vert \psi} 
      = \braket{x \vert \mleft( \sum_{p' \in \mathscr{P}} \ket{p'} \bra{p'} \mright) \vert \psi}     \\
    & = \sum_{p' \in \mathscr{P}} \braket { x \vert p' } \braket{ p' \vert \psi } = \sum_{p' \in \mathscr{P}} \braket { x \vert p' } \phi \mleft( p' \mright), 
\end{split} \\
%
\begin{split}\label{MomentumSpaceWaveFunctionsDiscr}
\phi \mleft( p \mright) 
    & = \braket{p \vert \psi} 
      = \braket{p \vert \hat{\mathbb{I}} \vert \psi} 
      = \braket{p \vert \mleft( \sum_{x' \in \mathscr{X}} \ket{x'} \bra{x'} \mright) \vert \psi}     \\
    & = \sum_{x' \in \mathscr{X}} \braket { p \vert x' } \braket{ x' \vert \psi } = \sum_{x' \in \mathscr{X}} \braket { p \vert x' } \psi \mleft( x' \mright) 
\end{split}
    \end{align}
\end{subequations}
\end{subequations}

\end{document}

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