How best to differentiate questions and answers in a document without compromising on typesetting and readability?

Question

Preliminary note: I don't think this question is completely or even to a large extent subjective, as it seeks how to achieve objective utilitarian goals. Please read:

I am solving my QFT assignments in LaTeX. They're basically just a succession of questions and answers and I would be interested in your opinions on how to best make it clear to the reader if text belongs to a question or gives an answer.

In the past, I've written the whole question in italics, as in

\newcommand{\question}[1]{\itshape #1}

However, I never found this to be satisfactory, as it makes the questions harder to read and the entire page layout seem quite disrupted. Suggestions?

Edit: Manuel asked for a compilable example, so here you go. I stripped pretty much every bit of code not essential off of the following. Note that the answer to b) is hard to discern from the list of questions.

\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}

Update: Two images in response to Peter Grill's solution.

Answer 1

Here is an example of using mdframed to highlight the answer portion:

References:

• You might want to try some of the other alternatives in Environments for visually setting text apart

Code:

\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}