[Tex/LaTex] Aligning a split up equation not at equal sign

alignequations

I currently have the following equation:

\begin{align}
\begin{split}
\left| \Psi \left( t \right) \right\rangle _{I}^{3}&=-\frac{1}{{{\hbar}^{2}}}\int_{0}^{t}{\hat{H}_{1}^{I}\left( t' \right)\,}\left( \int_{0}^{t}{\hat{H}_{1}^{I}\left( t'' \right)dt''} \right)dt' \notag \\ 
& =-\frac{1}{\hbar }\int_{0}^{t}{\hat{H}_{1}^{I}\left( t' \right)\,}\left( -\frac{g}{\Delta }\left( -{{{\hat{A}}}_{+}}\left( {{e}^{-i\Delta t'}}-1 \right)+{{{\hat{A}}}_{-}}\left( {{e}^{i\Delta t'}}-1 \right) \right) \right)dt' \notag \\ 
& =\frac{1}{\hbar }\int_{0}^{t}{\hbar g\left( {{{\hat{A}}}_{+}}{{e}^{-i\Delta t'}}+{{{\hat{A}}}_{-}}{{e}^{i\Delta t'}} \right)\,}\left( -\frac{g}{\Delta }\left( -{{{\hat{A}}}_{+}}\left( {{e}^{-i\Delta t'}}-1 \right)+{{{\hat{A}}}_{-}}\left( {{e}^{i\Delta t'}}-1 \right) \right) \right)dt' \notag \\
&=-\frac{{{g}^{2}}}{\Delta }\Bigg[ -\frac{1}{2i\Delta }{{e}^{-2i\Delta t'}}{{{\hat{A}}}_{+}}{{{\hat{A}}}_{+}}+\frac{1}{i\Delta }{{e}^{-i\Delta t'}}{{{\hat{A}}}_{+}}{{{\hat{A}}}_{+}}-{{{\hat{A}}}_{+}}{{{\hat{A}}}_{-}}-\frac{1}{i\Delta }{{e}^{-i\Delta t'}}{{{\hat{A}}}_{+}}{{{\hat{A}}}_{-}} \\ &+{{{\hat{A}}}_{-}}{{{\hat{A}}}_{+}}-\frac{1}{i\Delta }{{e}^{i\Delta t'}}{{{\hat{A}}}_{-}}{{{\hat{A}}}_{+}}-\frac{1}{2i\Delta }{{e}^{2i\Delta t'}}{{{\hat{A}}}_{-}}{{{\hat{A}}}_{+}}+\frac{1}{i\Delta }{{e}^{i\Delta t'}}{{{\hat{A}}}_{-}}{{{\hat{A}}}_{+}} \Bigg]_{0}^{t} \notag
\end{split}
\end{align}

From this I get 5 lines, where the last two are split up. But the last one is split up inside a bracket. My question is, can I align the last line, with the + sign, with the first minus sign, inside the bracket, instead of aligning with the = sign ?

Thanks in advance.

Best Answer

Please always post complete documents, not just fragments.

You can use \phantom

enter image description here

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align}
\begin{split}
\left| \Psi \left( t \right) \right\rangle _{I}^{3}&=-\frac{1}{{{\hbar}^{2}}}\int_{0}^{t}{\hat{H}_{1}^{I}\left( t' \right)\,}\left( \int_{0}^{t}{\hat{H}_{1}^{I}\left( t'' \right)dt''} \right)dt' \notag \\ 
& =-\frac{1}{\hbar }\int_{0}^{t}{\hat{H}_{1}^{I}\left( t' \right)\,}\left( -\frac{g}{\Delta }\left( -{{{\hat{A}}}_{+}}\left( {{e}^{-i\Delta t'}}-1 \right)+{{{\hat{A}}}_{-}}\left( {{e}^{i\Delta t'}}-1 \right) \right) \right)dt' \notag \\ 
& =\frac{1}{\hbar }\int_{0}^{t}{\hbar g\left( {{{\hat{A}}}_{+}}{{e}^{-i\Delta t'}}+{{{\hat{A}}}_{-}}{{e}^{i\Delta t'}} \right)\,}\left( -\frac{g}{\Delta }\left( -{{{\hat{A}}}_{+}}\left( {{e}^{-i\Delta t'}}-1 \right)+{{{\hat{A}}}_{-}}\left( {{e}^{i\Delta t'}}-1 \right) \right) \right)dt' \notag \\
&=-\frac{{{g}^{2}}}{\Delta }\Bigg[ -\frac{1}{2i\Delta }{{e}^{-2i\Delta t'}}{{{\hat{A}}}_{+}}{{{\hat{A}}}_{+}}+\frac{1}{i\Delta }{{e}^{-i\Delta t'}}{{{\hat{A}}}_{+}}{{{\hat{A}}}_{+}}-{{{\hat{A}}}_{+}}{{{\hat{A}}}_{-}}-\frac{1}{i\Delta }{{e}^{-i\Delta t'}}{{{\hat{A}}}_{+}}{{{\hat{A}}}_{-}} \\ 
&\phantom{{}=-\frac{{{g}^{2}}}{\Delta }\Bigg[}+{{{\hat{A}}}_{-}}{{{\hat{A}}}_{+}}-\frac{1}{i\Delta }{{e}^{i\Delta t'}}{{{\hat{A}}}_{-}}{{{\hat{A}}}_{+}}-\frac{1}{2i\Delta }{{e}^{2i\Delta t'}}{{{\hat{A}}}_{-}}{{{\hat{A}}}_{+}}+\frac{1}{i\Delta }{{e}^{i\Delta t'}}{{{\hat{A}}}_{-}}{{{\hat{A}}}_{+}} \Bigg]_{0}^{t} \notag
\end{split}
\end{align}
\end{document}

(I'm assuming this is cut down from some larger example, as there is no need to have an alignment just consisting of a split)

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