Underbrace outside parentheses

Question

How do I move the underbrace out of the big parentheses while still keeping it underneath the term?

\begin{alignat*}{4}
& \frac{E_{0-0}}{hc} &&= C+\frac{E'(\upsilon'=0)}{hc} &\\
&                    &&= C+\frac{1}{hc}\left(\hbar\omega'_e\left(0+\frac{1}{2}\right)-\underbrace{\hbar\chi'\omega'_e\left(0+\frac{1}{2}\right)^2}_{\approx0}\right) &  \quad\Big|\quad\parbox{9em}{\small{Vernachlässigung\\der Anharmonizität}}\\
&                    &&\approx C+\frac{\omega'_e}{4\pi c}
\end{alignat*}

Answer 1

I replaced alignat* with just align* as others also said and \Bigl,\Bigr are also better. The \frac{1}{2} looks unnecessarily big, so I used \tfrac. Finally, to lower the underbrace you can add a virtual rule like \vphantom{\rule[-1em]{0pt}{0pt}} noting that a \smash is required to keep line spacing consistent.

\documentclass{article}
\usepackage{amsmath}
\begin{document}

\begin{align*}
 \frac{E_{0-0}}{hc} &= C+\frac{E'(\upsilon'=0)}{hc} &\\
                    &= C+\frac{1}{hc}\Bigl(\hbar\omega'_e\left(0+\tfrac{1}{2}\right)-\smash{\underbrace{\vphantom{\rule[-1em]{0pt}{0pt}}\hbar\chi'\omega'_e\left(0+\tfrac{1}{2}\right)^2}_{\approx0}}\Bigr) &  \quad\Big|\quad\parbox{9em}{\small{Vernachlässigung\\der Anharmonizität}}\\
                    &\approx C+\frac{\omega'_e}{4\pi c}
\end{align*}

\end{document}

Answer 2

There are several aspects where this display can be improved.

The main one is avoiding \left and \right for those big parentheses. However there are more.

1. alignat is the wrong tool here, align is simpler and better

2. the \underbrace construction should be surrounded by braces, for avoiding issues with possible thin spaces inserted

3. you don't need to guess the width for the text, using a tabular is simpler, with the advantage that the vertical bar will be automatically adjusted

4. the exponent to the right parenthesis should be moved nearer to it

5. a thin space should be added after the underbraced part, so the brace will not clash with the parenthesis

6. since the underbrace ends up over nothing, we can \smash it

Here's the code.

\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[ngerman]{babel}

\usepackage{amsmath}

\begin{document}

\begin{align*}
\frac{E_{0-0}}{hc}
  &= C+\frac{E'(\upsilon'=0)}{hc} \\
  &= C+\frac{1}{hc}\biggl(
      \hbar\omega'_e\left(0+\frac{1}{2}\right)-
      \smash{\underbrace{\hbar\chi'\omega'_e\left(0+\frac{1}{2}\right)^{\!2}}_{\approx0}}
     \,\biggr)
     &&\mbox{\small
       \begin{tabular}{|l@{}}
       Vernachlässigung\\
       der Anharmonizität
       \end{tabular}%
     }\\
  &\approx C+\frac{\omega'_e}{4\pi c}
\end{align*}

\end{document}

Answer 3

One possibility:

\begin{align*}
 \frac{E_{0-0}}{hc} &= C+\frac{E'(\upsilon'=0)}{hc} \\
                    &= C+\frac{1}{hc}\biggl(\hbar\omega'_e\Bigl(0+\frac{1}{2}\Bigr)-\underbrace{\hbar\chi'\omega'_e\Bigl(0+\frac{1}{2}\Bigr)^{\!2}}_{\approx0}\,\biggr)
                    &  \quad\Big|\text{\small\begin{tabular}{l@{}}Vernachlässigung\\der Anharmonizität\end{tabular}}\\
                    &\approx C+\frac{\omega'_e}{4\pi c}
\end{align*}

Answer 4

To move it outside the right parenthesis, don't use the \left … \right pair and replace it with \biggl … \biggr. You'll have better sized parentheses and a better horizontal spacing.

In addition, note that with the alignat environment, the number of ampersands for n alignment groups is 2n-1: the first & is for the 1st alignment point, the 2nd introduces the second group, the 3rd & is for the alignment point of the second group, and so on.

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\begin{alignat*}{2}
  \frac{E_{0-0}}{hc} & = C+\frac{E'(\upsilon'=0)}{hc}\\
    & = C+\frac{1}{hc}\biggl(\hbar\omega'_e\Bigl(0+\frac{1}{2}\Bigr)-\underbrace{\hbar\chi'\omega'_e\Bigl(0+\frac{1}{2}\Bigr)²\biggr) \!\! }_{\approx0} & & \quad\Big|\quad\parbox{9em}{\small{Vernachlässigung \\der Anharmonizität}}\\
  &\approx C+\frac{\omega'_e}{4πc}
\end{alignat*}

\end{document}