先说明一下：我不认为这个问题完全是主观的，甚至在很大程度上不是主观的，因为它寻求的是如何实现客观的功利目标。请阅读：
我正在解决 QFT 作业LaTeX。它们基本上只是一系列的问题和答案，我很想知道您的意见，如何才能最好地让读者清楚地知道文本是属于问题还是给出了答案。
过去，我把整个问题都用斜体字写出来，比如
\newcommand{\question}[1]{\itshape #1}
但是，我始终不满意，因为它使问题更难阅读，并且整个页面布局看起来相当混乱。有什么建议吗？
编辑：Manuel 要求提供一个可编译的示例，所以您可以看看。我从以下内容中去掉了几乎所有不必要的代码。请注意，b) 的答案很难从问题列表中辨别出来。
\documentclass[12pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[english]{babel}
\usepackage{microtype}
\usepackage[a4paper,margin=25mm]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage[range-units = single]{siunitx}
\sisetup{detect-all,per-mode=fraction}
\usepackage[inline]{enumitem}
\setlist[enumerate]{leftmargin=0ex,label=\alph*)}
\setlist[itemize,1]{leftmargin=*}
\setlength{\parindent}{0ex}
\renewcommand{\vec}[1]{\boldsymbol{#1}}
\newcommand{\der}{\operatorname{d\!}{}}
\newcommand{\spint}[1][x]{\int_{\mathbb{R}^3} \! \der^3#1 \,}
\newcommand{\moint}{\int_{\mathbb{R}^3} \! \frac{\der^3p}{\(\num{2}\pi\)^{\num{3}}} \,}
\newcommand{\vnabla}{\vec{\nabla}}
\renewcommand{\(}{\left(}
\renewcommand{\)}{\right)}
\renewcommand{\[}{\left[}
\renewcommand{\]}{\right]}
\let\phi\varphi
\begin{document}
\section{Canonical commutation relations}
Starting from the mode expansion of the Schrödinger fields
\begin{align}
&\phi\(\vec{x}\) = \moint \frac{\num{1}}{\sqrt{\num{2}\omega_{\vec{p}}}} \(a\(\vec{p}\)e^{i\vec{p}\vec{x}} + a^\dagger\(\vec{p}\)e^{-i\vec{p}\vec{x}}\), \\
\text{and} \quad &\pi\(\vec{x}\) = \moint (-i) \sqrt{\frac{\omega_{\vec{p}}}{\num{2}}} \(a\(\vec{p}\)e^{i\vec{p}\vec{x}} - a^\dagger\(\vec{p}\)e^{-i\vec{p}\vec{x}}\),
\end{align}
we’ll derive the canonical commutation relations for the modes $a\(\vec{p}\)$ and $a^\dagger\(\vec{p}\)$ using the steps below.
\begin{enumerate}
\item Perform a Fourier transformation to deduce the relation between the Fourier modes $\tilde{\phi}\(\vec{p}\)$, $\tilde{\pi}\(\vec{p}\)$, and the modes $a\(\vec{p}\)$ and $a^\dagger\(\vec{p}\)$.
\item Derive the commutation relations
\begin{equation}
\[\tilde{\phi}\(\vec{p}\),\tilde{\pi}\(\vec{q}\)\] = i \(\num{2}\pi\)^{\num{3}} \delta^{\(\num{3}\)}\(\vec{p} + \vec{q}\)
\end{equation}
from the canonical commutation relations for $\phi\(\vec{x}\)$ and $\pi\(\vec{x}\)$.
%%%%%% All text belongs to questions except for this block. %%%%%%
The canonical commutation relations for $\phi\(\vec{x}\)$ and $\pi\(\vec{x}\)$ states that $\[\phi\(\vec{x}\),\pi\(\vec{y}\)\] = i \delta^{\(\num{3}\)}\(\vec{x} - \vec{y}\)$. Therefore
\begin{equation}
\begin{aligned}
\[\tilde{\phi}\(\vec{p}\),\tilde{\pi}\(\vec{q}\)\]
&= \spint\spint[y] e^{-i\vec{p}\vec{x}} e^{-i\vec{p}\vec{y}} \underbrace{\[\phi\(\vec{x}\),\pi\(\vec{y}\)\]}_{i \delta^{\(\num{3}\)}\(\vec{x} - \vec{y}\)} \\
&= i \spint e^{-i\(\vec{p}+\vec{q}\)\vec{x}}
= i \(\num{2}\pi\)^{\num{3}} \delta^{\(\num{3}\)}\(\vec{p} + \vec{q}\)
\end{aligned}
\end{equation}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item Conclude that
\begin{equation}
\[a\(\vec{p}\),a^\dagger\(\vec{q}\)\] = \(\num{2}\pi\)^{\num{3}} \delta^{\(\num{3}\)}\(\vec{p} - \vec{q}\)
\quad \text{and} \quad
\[a\(\vec{p}\),a\(\vec{q}\)\] = \[a^\dagger\(\vec{p}\),a^\dagger\(\vec{q}\)\] = \num{0}.
\end{equation}
\end{enumerate}
\end{document}
更新：两张图片响应 Peter Grill 的解决方案。
答案1
mdframed以下是用于突出显示答案部分的示例：
参考：
- 你可能想尝试一些其他的替代方案用于在视觉上区分文本的环境
代码：
\documentclass[12pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[english]{babel}
\usepackage{microtype}
\usepackage[a4paper,margin=25mm]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage[range-units = single]{siunitx}
\sisetup{detect-all,per-mode=fraction}
\usepackage[inline]{enumitem}
\setlist[enumerate]{leftmargin=0ex,label=\alph*)}
\setlist[itemize,1]{leftmargin=*}
\setlength{\parindent}{0ex}
\renewcommand{\vec}[1]{\boldsymbol{#1}}
\newcommand{\der}{\operatorname{d\!}{}}
\newcommand{\spint}[1][x]{\int_{\mathbb{R}^3} \! \der^3#1 \,}
\newcommand{\moint}{\int_{\mathbb{R}^3} \! \frac{\der^3p}{\(\num{2}\pi\)^{\num{3}}} \,}
\newcommand{\vnabla}{\vec{\nabla}}
\renewcommand{\(}{\left(}
\renewcommand{\)}{\right)}
\renewcommand{\[}{\left[}
\renewcommand{\]}{\right]}
\let\phi\varphi
\usepackage{xcolor}
\usepackage{mdframed}
\newmdenv[%
leftmargin=-5pt,
rightmargin=-5pt,
innerleftmargin=5pt,
innerrightmargin=5pt,
backgroundcolor=brown!10,
]{Answer}%
\begin{document}
\section{Canonical commutation relations}
Starting from the mode expansion of the Schrödinger fields
\begin{align}
&\phi\(\vec{x}\) = \moint \frac{\num{1}}{\sqrt{\num{2}\omega_{\vec{p}}}} \(a\(\vec{p}\)e^{i\vec{p}\vec{x}} + a^\dagger\(\vec{p}\)e^{-i\vec{p}\vec{x}}\), \\
\text{and} \quad &\pi\(\vec{x}\) = \moint (-i) \sqrt{\frac{\omega_{\vec{p}}}{\num{2}}} \(a\(\vec{p}\)e^{i\vec{p}\vec{x}} - a^\dagger\(\vec{p}\)e^{-i\vec{p}\vec{x}}\),
\end{align}
we’ll derive the canonical commutation relations for the modes $a\(\vec{p}\)$ and $a^\dagger\(\vec{p}\)$ using the steps below.
\begin{enumerate}
\item Perform a Fourier transformation to deduce the relation between the Fourier modes $\tilde{\phi}\(\vec{p}\)$, $\tilde{\pi}\(\vec{p}\)$, and the modes $a\(\vec{p}\)$ and $a^\dagger\(\vec{p}\)$.
\item Derive the commutation relations
\begin{equation}
\[\tilde{\phi}\(\vec{p}\),\tilde{\pi}\(\vec{q}\)\] = i \(\num{2}\pi\)^{\num{3}} \delta^{\(\num{3}\)}\(\vec{p} + \vec{q}\)
\end{equation}
from the canonical commutation relations for $\phi\(\vec{x}\)$ and $\pi\(\vec{x}\)$.
\begin{Answer}
%%%%%% All text belongs to questions except for this block. %%%%%%
The canonical commutation relations for $\phi\(\vec{x}\)$ and $\pi\(\vec{x}\)$ states that $\[\phi\(\vec{x}\),\pi\(\vec{y}\)\] = i \delta^{\(\num{3}\)}\(\vec{x} - \vec{y}\)$. Therefore
\begin{equation}
\begin{aligned}
\[\tilde{\phi}\(\vec{p}\),\tilde{\pi}\(\vec{q}\)\]
&= \spint\spint[y] e^{-i\vec{p}\vec{x}} e^{-i\vec{p}\vec{y}} \underbrace{\[\phi\(\vec{x}\),\pi\(\vec{y}\)\]}_{i \delta^{\(\num{3}\)}\(\vec{x} - \vec{y}\)} \\
&= i \spint e^{-i\(\vec{p}+\vec{q}\)\vec{x}}
= i \(\num{2}\pi\)^{\num{3}} \delta^{\(\num{3}\)}\(\vec{p} + \vec{q}\)
\end{aligned}
\end{equation}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{Answer}
\item Conclude that
\begin{equation}
\[a\(\vec{p}\),a^\dagger\(\vec{q}\)\] = \(\num{2}\pi\)^{\num{3}} \delta^{\(\num{3}\)}\(\vec{p} - \vec{q}\)
\quad \text{and} \quad
\[a\(\vec{p}\),a\(\vec{q}\)\] = \[a^\dagger\(\vec{p}\),a^\dagger\(\vec{q}\)\] = \num{0}.
\end{equation}
\end{enumerate}
\end{document}