\hfill in math mode for multiple alignments

Question

I would like to reproduce something like the screenshot attached here, with two separate equations in each line, each set of equations aligned amongst themselves.

On my end, LaTeX always pushes everything left of the second alignment symbol to the farthest left it can be, as if it is all part of the first equation, while I would like the left hand side of the second equation in the same line be immediately next to its own equal sign.

Specifically for the case attached here, I would like to align the two equations respectively with $\to$ and $=$ separately.

Also, if possible, ideally I would like to implement this with just the align environment.

Minimum not-working code:

\documentclass[11pt, oneside]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[english]{babel}
\usepackage{amsmath}
\begin{document}
\begin{align}
\left(\begin{array}{c}{A_{\mu}} \\ {\rho_{\mu}^{*}}\end{array}\right) &\rightarrow\left(\begin{array}{cc}{\cos \theta} & {-\sin \theta} \\ {\sin \theta} & {\cos \theta}\end{array}\right)\left(\begin{array}{c}{A_{\mu}} \\ {\rho_{\mu}^{*}}\end{array}\right), \qquad\qquad\qquad\qquad\tan \theta&=\frac{g_{e l}}{g_{*}}\\
\left(\begin{array}{c}{\psi_{L}} \\ {\chi_{L}}\end{array}\right) &\rightarrow\left(\begin{array}{cc}{\cos \varphi_{\psi_{L}}} & {-\sin \varphi_{\psi_{L}}} \\ {\sin \varphi_{\psi_{L}}} & {\cos \varphi_{\psi_{L}}}\end{array}\right)\left(\begin{array}{c}{\psi_{L}} \\ {\chi_{L}}\end{array}\right), \hfill \tan \varphi_{\psi_{L}}&=\frac{\Delta}{m}\\
\left(\begin{array}{c}{\tilde{\psi}_{R}} \\ {\tilde{\chi}_{R}}\end{array}\right) &\rightarrow\left(\begin{array}{cc}{\cos \varphi_{\tilde{\psi}_{R}}} & {-\sin \varphi_{\tilde{\psi}_{R}}} \\ {\sin \varphi_{\tilde{\psi}_{R}}} & {\cos \varphi_{\tilde{\psi}_{R}}}\end{array}\right)\left(\begin{array}{c}{\tilde{\psi}_{R}} \\ {\tilde{\chi}_{R}}\end{array}\right), \quad \tan \varphi_{\tilde{\psi}_{R}}&=\frac{\tilde{\Delta}}{\tilde{m}}
\end{align}
\end{document}

Answer 1

like this?

with use of pmatrix from asmath and extend align environment with one more ampersand:

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

\begin{document}
\begin{align}
\begin{pmatrix}
    A_{\mu}\\ \rho_{\mu}^{*}
\end{pmatrix}
    & \rightarrow   \begin{pmatrix}
                    \cos \theta & -\sin \theta \\
                    \sin \theta &  \cos \theta
                    \end{pmatrix}\begin{pmatrix}
                                    A_{\mu} \\ \rho_{\mu}^{*}
                                 \end{pmatrix},
    &   \tan \theta & = \frac{g_{e l}}{g_{*}}         \\
\begin{pmatrix}\psi_{L} \\ \chi_{L} \end{pmatrix}
    & \rightarrow   \begin{pmatrix}
                    \cos \varphi_{\psi_{L}} & -\sin \varphi_{\psi_{L}} \\ 
                    \sin \varphi_{\psi_{L}} &  \cos \varphi_{\psi_{L}}
                    \end{pmatrix}\begin{pmatrix}
                                    \psi_{L} \\ \chi_{L}
                                 \end{pmatrix},
    &   \tan \varphi_{\psi_{L}} 
            & = \frac{\Delta}{m}    \\
\begin{pmatrix}
    \tilde{\psi}_{R} \\  \tilde{\chi}_{R}
\end{pmatrix}
    & \rightarrow   \begin{pmatrix}
                    \cos \varphi_{\tilde{\psi}_{R}} & -\sin \varphi_{\tilde{\psi}_{R}}  \\ 
                    \sin \varphi_{\tilde{\psi}_{R}} &  \cos \varphi_{\tilde{\psi}_{R}}
                    \end{pmatrix}\begin{pmatrix}
                                    \tilde{\psi}_{R} \\\tilde{\chi}_{R}
                                 \end{pmatrix},
    &   \tan \varphi_{\tilde{\psi}_{R}} 
            & =\frac{\tilde{\Delta}}{\tilde{m}}
\end{align}
\end{document}    

Answer 2

I'd propose alignat, that gives more control of spacing between the column pairs.

The second parts of each row should not, in my opinion, be aligned at the equals sign, because the prominence should be to “tan”.

\documentclass[11pt, oneside]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[english]{babel}
\usepackage{amsmath}
\begin{document}
\begin{alignat}{2}
\begin{pmatrix} A_{\mu} \\ \rho_{\mu}^{*} \end{pmatrix}
  & \rightarrow
    \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}
    \begin{pmatrix} A_{\mu} \\ \rho_{\mu}^{*} \end{pmatrix},
  &\quad& \tan \theta = \frac{g_{e l}}{g_{*}}
\\
\begin{pmatrix} \psi_{L} \\ \chi_{L} \end{pmatrix}
  & \rightarrow
    \begin{pmatrix}
      \cos \varphi_{\psi_{L}} & -\sin \varphi_{\psi_{L}} \\
      \sin \varphi_{\psi_{L}} & \cos \varphi_{\psi_{L}}
    \end{pmatrix}
    \begin{pmatrix} \psi_{L} \\ \chi_{L} \end{pmatrix},
  && \tan \varphi_{\psi_{L}} = \frac{\Delta}{m}
\\
\begin{pmatrix} \tilde{\psi}_{R} \\ \tilde{\chi}_{R} \end{pmatrix}
   & \rightarrow
     \begin{pmatrix}
       \cos \varphi_{\tilde{\psi}_{R}} & -\sin \varphi_{\tilde{\psi}_{R}} \\
       \sin \varphi_{\tilde{\psi}_{R}} & \cos \varphi_{\tilde{\psi}_{R}}
     \end{pmatrix}
     \begin{pmatrix} \tilde{\psi}_{R} \\ \tilde{\chi}_{R} \end{pmatrix},
   && \tan \varphi_{\tilde{\psi}_{R}} = \frac{\tilde{\Delta}}{\tilde{m}}
\end{alignat}

\end{document}

Your opinion may differ. In any case, avoid overusing braces.