Curly brace to insert something into an equation? Like an inverted underbrace

Question

I have a generic equation, into which I want to "insert" something so that the reader can see what I inserted into the equation and at which point.

E.g.

\documentclass{plain}

\begin{document}
\[ f = < x | R y > \]

Insert $\int d^3 p |p> <p|$ = 1 before y.
\end{document}

What I imagine is a horizontal curly brace, which extends underneath the equation, and which has its tip touching a spot just slightly to the left of the $y$, while in the brace there is written $\int d^3 p |p> <p|$. It would look something like an inverted \underbrace{foo}_{bar}, but I have no idea how to achieve something like that.

In ASCII art this would maybe look like this:

f = <x | R y >            (1)
        __|__
   ____|     |____
  | Insert here! | 

Answer 1

Ok, before people wonder what happened to the tikz solution, here is an alternate method, using, well... tikz. \tikzmark is used to name the location in the equation that you want to refer to. This name is then given to \Insert along with various options to control the behavior of the brace and the line connecting them.

The parameters passed to \Insert are:

#1. Vertical shift to be applied to the brace

#2. Options to control the behavior of the connecting line.

#3. Name of the marker in the equation.

#4. Text to be braced, which is to be inserted.

Notes:

• This does require two runs. First one to determine the locations, and the second to do the drawing.
• Probably will have issues if the text begin inserted is not on the first line as the arrow will probably overlap other text.

References:

• The brace is adapted from adding a large brace next to a body of text

Code:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\usetikzlibrary{decorations.pathreplacing}
\newcommand{\tikzmark}[1]{\tikz[overlay,remember picture] \node (#1) {};}
\newcommand{\BraceAmplitude}{0.5em}%
\newcommand{\Insert}[5][0.0ex]{%
    \tikzmark{a}#5\tikzmark{b}%
    \begin{tikzpicture}[overlay,remember picture]
\draw [decoration={brace,amplitude=\BraceAmplitude},decorate,ultra thick,blue,#2]
        ($(a.north)+(0,#1)$) -- ($(b.north)+(0,#1)$);
\draw [thick, blue, #3] ($(a.north)!0.5!(b.north)+(0,#1)+(0,\BraceAmplitude+2pt)$) to (#4);
    \end{tikzpicture}%
}
\begin{document}
[ f = < x | R \tikzmark{MyMarker} y > ]
\bigskip
In the above equation, we need to insert \Insert[1.5ex]{orange}{out=90, in=-90,->,shorten >=-1pt}{MyMarker}{$\int d^3 p |p> <p| = 1$} before $y$.
[ f = < x | R \tikzmark{MyMarkerA} y \tikzmark{MyMarkerB}> ]
\bigskip
Insert \Insert[1.5ex]{blue}{out=90, in=-90,->,shorten >=-1pt,red}{MyMarkerA}{$\int d^3 p |p> <p| = 1$} before $y$, and some more
\Insert[1.0ex]{brown}{out=135, in=-70,->,shorten >=-2pt}{MyMarkerB}{\textbf{stuff}} after $y$.
\end{document}

Answer 2

Here's one possible solution (in LaTeX2e) using an \overbrace and \mathclap (from the mathtools package); the vertical position was adjusted using \raisebox:

\documentclass{article}
\usepackage{mathtools}

\begin{document}

\[
  f = < x | R\mathclap{\raisebox{-4ex}{$\overbrace{\textstyle\int d^3 p |p> <p|}$}} y > 
\]

\end{document}

I also would suggest you to use \langle, \rangle instead of the relational symbols <, >a:

\documentclass{report}
\usepackage{mathtools}

\begin{document}

\[
  f = \langle x | R\mathclap{\raisebox{-4ex}{$\overbrace{\textstyle\int d^3 p |p\rangle \langle p|}$}} y \rangle 
\]

\end{document}

Answer 3

Here's a variant of Gonzalo's code that uses \mathop{}\limits instead of \raisebox. The output is almost the same.

\documentclass{article}
\usepackage{mathtools}
\begin{document}
\[
  f = \langle x | R\mathclap{\mathop{}\limits_{\overbrace{
                   \textstyle\int d^3 p |p\rangle \langle p|}}}y
      \rangle 
\]
\end{document}

---

If you want something more flexible, use the \braceinsert macro from the code below. It takes one optional argument (the amount by which the brace should be lowered, with default 0ex) and one mandatory argument (the stuff you want under the brace). Moreover, it takes care of the case that the \braceinsert is surrounded by \left...\right delimiters, like this:

To be precise, you'll have to use \bileft...\biright instead of \left...\right (like braceinsert left and right). After the outermost \biright, only explicit superscripts will work, so instead of ' you'd have to use ^\prime. (Subscripts won't work properly, but I don't know when one would want one.)

\documentclass{article}
\usepackage{mathtools}
\setlength{\textwidth}{11cm}
\makeatletter
\newcounter{left@right}
\newcommand*\bileft[1]{\left#1\stepcounter{left@right}}
\newcommand*\biright[1]{\right#1%
    \@ifnextchar^{\with@superscript}{\without@superscript}}
\def\with@superscript^#1{^{#1}\without@superscript}
\def\without@superscript{%
    \addtocounter{left@right}{-1}%
    \ifnum\theleft@right=0
      \vphantom{\brace@insert@strut}
      \gdef\brace@insert@strut{}
    \fi
    }
\def\brace@insert@strut{}
\newcommand{\braceinsert}[2][0ex]{%
    \def\insert@material{%
        \mathop{\rule[-#1]{0pt}{0pt}}\limits_{\overbrace{#2}}%
        }%
    \expandafter\g@addto@macro\expandafter\brace@insert@strut
        \expandafter{\insert@material}
    \ifnum\theleft@right>0
      \smash{\mathclap{\insert@material}}
    \else
      \mathclap{\insert@material}
    \fi
    }
\makeatother
\begin{document}
In the formula
\[
  f = 5 \bileft( b^2 + \langle x | R
      \braceinsert[0.2ex]{\textstyle\int d^3 p |p\rangle \langle p|}
      y \rangle \biright)^2,
\]
the depth of the stuff under the brace is added only \emph{after}
\verb|\biright)^2|.
\end{document}

^{(Maybe this is wayyy too complicated?)}