Missing bracket

Question

I'm having a hard time to figure out what is wrong with the follwoing MWE

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align*}
\langle n,l',l_s',m',m_s'\vert H_{\rm SO}\vert n,l,l_s,m,m_s \rangle = \frac{1}{2m^2c^2 a_0^3n^3}\frac{\delta_{l'}^l}{l(2l+1)(l+1)}
\left(
& \sqrt{l(l+1)-m(m+1)}\sqrt{3/4-m_s(m_s-1)}\delta_{m'}^{m+1}\delta_{m_s'}^{m_s-1}\\
& \sqrt{l(l+1)-m(m-1)}\sqrt{3/4-m_s(m_s+1)}\delta_{m'}^{m-1}\delta_{m_s'}^{m_s+1}\\
& (m+m_s)\delta_m^{m'}\delta_{m_s}^{m_s'}
\right)
\end{align*}
\end{document}

It says that I forgot to close a bracket, but I fail to see where...

Raven · Answer 1 · 2018-12-19T16:23:22.243

It's not possible to use \left( and \right) across line breaks. Replacing those with normal braces should do the trick.

You can use a workaround as described here. In your case this would be

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align*}
&\langle n,l',l_s',m',m_s'\vert H_{\rm SO}\vert n,l,l_s,m,m_s \rangle\\
&= \frac{1}{2m^2c^2 a_0^3n^3}\frac{\delta_{l'}^l}{l(2l+1)(l+1)} \left( \sqrt{l(l+1)-m(m+1)}\sqrt{3/4-m_s(m_s-1)}\delta_{m'}^{m+1}\delta_{m_s'}^{m_s-1}\right.\\
& \left. \sqrt{l(l+1)-m(m-1)}\sqrt{3/4-m_s(m_s+1)}\delta_{m'}^{m-1}\delta_{m_s'}^{m_s+1}  (m+m_s)\delta_m^{m'}\delta_{m_s}^{m_s'}\right)
\end{align*}
\end{document}

score 0 · Answer 2 · answered Dec 19 '18 at 16:22

As pointed by Raven, you can simply substitute your MWE with:

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align*}
\langle n,l',l_s',m',m_s'\vert H_{\rm SO}\vert n,l,l_s,m,m_s \rangle &= \frac{1}{2m^2c^2 a_0^3n^3}\frac{\delta_{l'}^l}{l(2l+1)(l+1)}
\left( \sqrt{l(l+1)-m(m+1)}\right.\\
&\sqrt{3/4-m_s(m_s-1)}\delta_{m'}^{m+1}\delta_{m_s'}^{m_s-1}
 \sqrt{l(l+1)-m(m-1)}\\
&\left.\sqrt{3/4-m_s(m_s+1)} \delta_{m'}^{m-1}\delta_{m_s'}^{m_s+1}(m+m_s)\delta_m^{m'}\delta_{m_s}^{m_s'}
\right)
\end{align*}
\end{document}

to get

Missing bracket

2 Answers2