1

My working example is:

\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}

It produces the following output: enter image description here *I've capitalized the F in Fourier transformation after taking the screenshot.

My issue is really with equations 0.2. I want them to be nested in the way they are, but they obviously don't fit, but I think breaking the up each would make this into a real monster. Any suggestion on how I could just really save space just in this section of my overall document?

1

I would just keep the aligns and split things in two lines such that the perhaps most important steps are directly below each other. I also recommend using upright differentials, and put spaces before the punctuations. Too many unnecessary mistakes have happened because someone mistook an $x_i,$ as an $x_{i'}$. I am not convinced by the nested subequations but you seem to like it.

\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}

enter image description here

| improve this answer | |
  • thank you for pointing out the negative spacing after the integral sign and upright differential as well! – Markus Gratis May 22 at 14:08
1

Like this?

enter image description here

With using 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}
| improve this answer | |
  • I like Schrödinger's cat's solution more tbh, but thank you! – Markus Gratis May 22 at 13:58

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