# Typesetting d'Alembertian symbol

I've seen that usually, the d'Alembertian is written using the command \Box, however, this displays a square with all sides identical. I would like to write it in this other way:

in which, the right and below sides are bolder, so that resembles more to the Laplacian symbol.

• Thanks for the welcome! \square is indeed a better option since the scaling is better to the \Box, however, it still presents no difference in the strokes. – Janstillerion Nov 9 '18 at 17:11

Build a square with thicker lines the same size of \square:

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

\makeatletter
\newcommand{\dalembertian}{\mathop{\mathpalette\dalembertian@\relax}}
\newcommand{\dalembertian@}[2]{%
\begingroup
\sbox\z@{$\m@th#1\square$}%
\dimen0=\fontdimen8
\ifx#1\displaystyle\textfont\else
\ifx#1\textstyle\textfont\else
\ifx#1\scriptstyle\scriptfont\else
\scriptscriptfont\fi\fi\fi3
\makebox[\wd\z@]{%
\hbox to \ht\z@{%
\vrule width \dimen0
\kern-\dimen0
\vbox to \ht\z@{
\hrule height \dimen0 width \ht\z@
\vss
\hrule height 2\dimen0
}%
\kern-2.5\dimen0
\vrule width 2.5\dimen0
}%
}%
\endgroup
}
\makeatother

\begin{document}

$\dalembertian = \mathop{{}\Delta}-\partial_0^2$

$\scriptstyle\dalembertian = \mathop{{}\Delta}-\partial_0^2$

$\scriptscriptstyle\dalembertian = \mathop{{}\Delta}-\partial_0^2$

{\Large$\dalembertian = \mathop{{}\Delta}-\partial_0^2$}

\end{document}


• Thanks, this is exactly what I was looking for. – Janstillerion Nov 12 '18 at 17:13

To increase the boarder on two sides, you could use the \ding{113} symbol from pifont

\documentclass{article}

\usepackage{pifont}
\usepackage{amsmath}

\begin{document}

$\text{\ding{113}} \Delta$

\end{document}


Just uses \rules. Then, \scalerel to make it match the height of \Delta.

\documentclass{article}
\usepackage{stackengine,scalerel}
\newcommand\dAlaux{%
\Shortstack{\rule{12pt}{.6pt}\\
\rule{.6pt}{10pt}\kern10pt\rule{1.4pt}{10pt}\\
\rule{12pt}{1.4pt}}%
}
\newcommand\dAl{%
\setstackgap{S}{0pt}%
\setstackEOL{\\}%
\scalerel*{\kern1pt\dAlaux\kern1pt}{\Delta}%
}
\begin{document}
$\dAl = \Delta - \partial_0^2,\quad \phi\dAl\phi e^{\phi\dAl\phi}$
\end{document}


A slightly thicker version, by adjusting the \rule dimensions...

\documentclass{article}
\usepackage{stackengine,scalerel}
\newcommand\dAlaux{%
\Shortstack{\rule{11pt}{.7pt}\\
\rule{.7pt}{9pt}\kern9pt\rule{1.3pt}{9pt}\\
\rule{11pt}{1.35pt}}%
}
\newcommand\dAl{%
\setstackgap{S}{0pt}%
\setstackEOL{\\}%
\scalerel*{\kern1pt\dAlaux\kern1pt}{\Delta}%
}
\begin{document}
$\dAl = \Delta - \partial_0^2,\quad \phi\dAl\phi e^{\phi\dAl\phi}$
\end{document}


Here's a version using scalebox and overlapping three copies of \square so that the thickness can be tweaked. The thickness can be changed by increasing or reducing the second argument to \add@dalemb.

For instance, with 03 and 06:

and with 05 and 10:

Thicker than this, another \add@dalemb would be needed to avoid white spaces between the \squares.

\documentclass[a4paper,12pt]{article}

\usepackage{amssymb}
\usepackage{graphicx}

\makeatletter
\DeclareRobustCommand\dalemb{\mathpalette\inner@dalemb{}}
\def\inner@dalemb#1{%
\square
}
\sbox0{\scalebox{1.#2}{$#1\square$}}%
\rlap{\lower0.#2\ht0\box0}%
}
\makeatother

\begin{document}

$\Delta\dalemb$
$\partial^{\Delta\dalemb}$
$\partial^{\partial^{\Delta\dalemb}}$

\end{document}