3

First of all, I wish to thank you all for your past contributions on some questions I made - they were invaluable indeed!

Presently, I come here with yet another question: How to properly align a table containing matrices?

The code's below! Thanks in advance!

\documentclass{amsart}
\begin{document}

\begin{table}[]
\centering
\caption{Algebraic expressions of some 3-j symbols}
\label{my-label}
    \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} &=& \begin{pmatrix} j_2 & j_3 & j_1 \\ m_2 & m_3 & m_1 \end{pmatrix} = \begin{pmatrix} j_3 & j_1 & j_2 \\ m_3 & m_1 & m_2 \end{pmatrix} \\
    \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} &=& (-1)^{{j_1}+{j_2}+{j_3}} \begin{pmatrix}j_2 & j_1 & j_3 \\ m_2 & m_1 & m_3 \end{pmatrix} \\
    \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} &=& (-1)^{{j_1}+{j_2}+{j_3}} \begin{pmatrix}j_1 & j_2 & j_3 \\ -m_1 & -m_2 & -m_3 \end{pmatrix} \\
    \begin{pmatrix} j+1 & j & 1     \\ m & -m-1 & 1    \end{pmatrix} &=& (-1)^{{j-m+1}}{\left[\frac{(j-m)(j-m+1)}{(2j+3)(2j+2)(2j+1)}\right]}^{\frac{1}{2}} \\
    \begin{pmatrix} j+1 & j & 1     \\ m & -m & 1      \end{pmatrix} &=& (-1)^{{j-m-1}}{\left[\frac{(j-m+1)(j-m+1)}{(2j+3)(2j+2)(2j+1)}\right]}^{\frac{1}{2}} \\
    \begin{pmatrix} j   & j & 1     \\ m & -m-1 & 1    \end{pmatrix} &=& (-1)^{{j-m}}{\left[\frac{(j-m)(j-m+1)}{j(2j+1)(2j+2)}\right]}^{\frac{1}{2}}  \\
    \begin{pmatrix} j   & j & 1     \\ m & -m & 0      \end{pmatrix} &=& (-1)^{{j-m}}{\left[\frac{m^2}{j(j+1)(2j+1)}\right]}^{\frac{1}{2}}
 \end{table}

 \end{document}

Current output: enter image description here

2
  • As a reminder, when you ask a question you should include a "Minimal Working Example" (MWE) that starts with \documentclass, includes any \usepackage commands, and ends with \end{document}. The MWE should compile without errors, even if it does not produce the output you desire. Then you are much more likely to get help on this site. If you are having trouble getting started, you should first learn the basics of matrices and tables. Many resources are available online. I suggest Google "latex tables" to start.
    – Sandy G
    Nov 3 '17 at 19:49
  • @SandyG - thanks a lot for the editing and for kindly pointing me the right way to do things around here!
    – Strelok
    Nov 6 '17 at 10:44
5

There's no rule preventing usage of align* in a table:

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{table}

\caption{Algebraic expressions of some 3-j symbols}
\label{my-label}

\begin{align*}
\begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix}
  &= \begin{pmatrix} j_2 & j_3 & j_1 \\ m_2 & m_3 & m_1 \end{pmatrix}
   = \begin{pmatrix} j_3 & j_1 & j_2 \\ m_3 & m_1 & m_2 \end{pmatrix} \\
\begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix}
  &= (-1)^{{j_1}+{j_2}+{j_3}}
     \begin{pmatrix}j_2 & j_1 & j_3 \\ m_2 & m_1 & m_3 \end{pmatrix} \\
\begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix}
  &= (-1)^{{j_1}+{j_2}+{j_3}}
     \begin{pmatrix}j_1 & j_2 & j_3 \\ -m_1 & -m_2 & -m_3 \end{pmatrix} \\
\begin{pmatrix} j+1 & j & 1     \\ m & -m-1 & 1    \end{pmatrix}
  &= (-1)^{{j-m+1}}{\left[\frac{(j-m)(j-m+1)}{(2j+3)(2j+2)(2j+1)}\right]}^{\frac{1}{2}} \\
\begin{pmatrix} j+1 & j & 1     \\ m & -m & 1      \end{pmatrix}
  &= (-1)^{{j-m-1}}{\left[\frac{(j-m+1)(j-m+1)}{(2j+3)(2j+2)(2j+1)}\right]}^{\frac{1}{2}} \\
\begin{pmatrix} j   & j & 1     \\ m & -m-1 & 1    \end{pmatrix}
  &= (-1)^{{j-m}}{\left[\frac{(j-m)(j-m+1)}{j(2j+1)(2j+2)}\right]}^{\frac{1}{2}} \\
\begin{pmatrix} j   & j & 1     \\ m & -m & 0      \end{pmatrix}
  &= (-1)^{{j-m}}{\left[\frac{m^2}{j(j+1)(2j+1)}\right]}^{\frac{1}{2}}
\end{align*}

\end{table} 

\end{document}

enter image description here

1
  • Simple and elegant!
    – Strelok
    Nov 6 '17 at 10:46
2

I am not quite sure what you mean by properly aligned, but you can align the equal signs by using the tabular environment.

Code:

\documentclass{amsart}
\begin{document}

\begin{table}[]
\centering
\caption{Algebraic expressions of some 3-j symbols}
\label{my-label}

\begin{tabular}{rcl}

    \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} &=& \begin{pmatrix} j_2 & j_3 & j_1 \\ m_2 & m_3 & m_1 \end{pmatrix} = \begin{pmatrix} j_3 & j_1 & j_2 \\ m_3 & m_1 & m_2 \end{pmatrix} \\ 

    \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} &=& (-1)^{{j_1}+{j_2}+{j_3}} \begin{pmatrix}j_2 & j_1 & j_3 \\ m_2 & m_1 & m_3 \end{pmatrix} \\

    \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} &=& (-1)^{{j_1}+{j_2}+{j_3}} \begin{pmatrix}j_1 & j_2 & j_3 \\ -m_1 & -m_2 & -m_3 \end{pmatrix} \\

    \begin{pmatrix} j+1 & j & 1     \\ m & -m-1 & 1    \end{pmatrix} &=& (-1)^{{j-m+1}}{\left[\frac{(j-m)(j-m+1)}{(2j+3)(2j+2)(2j+1)}\right]}^{\frac{1}{2}} \\

    \begin{pmatrix} j+1 & j & 1     \\ m & -m & 1      \end{pmatrix} &=& (-1)^{{j-m-1}}{\left[\frac{(j-m+1)(j-m+1)}{(2j+3)(2j+2)(2j+1)}\right]}^{\frac{1}{2}} \\

    \begin{pmatrix} j   & j & 1     \\ m & -m-1 & 1    \end{pmatrix} &=& (-1)^{{j-m}}{\left[\frac{(j-m)(j-m+1)}{j(2j+1)(2j+2)}\right]}^{\frac{1}{2}}  \\

    \begin{pmatrix} j   & j & 1     \\ m & -m & 0      \end{pmatrix} &=& (-1)^{{j-m}}{\left[\frac{m^2}{j(j+1)(2j+1)}\right]}^{\frac{1}{2}}\\

\end{tabular}

\end{table}
\end{document}

Which yields: enter image description here

Personally, I think the rows seem a bit cramped (vertically) so you could widen them by doing something like this:

\documentclass{amsart}
\begin{document}

\begin{table}[]
\centering
\caption{Algebraic expressions of some 3-j symbols}
\label{my-label}

\begin{tabular}{rcl}

    \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} &=& \begin{pmatrix} j_2 & j_3 & j_1 \\ m_2 & m_3 & m_1 \end{pmatrix} = \begin{pmatrix} j_3 & j_1 & j_2 \\ m_3 & m_1 & m_2 \end{pmatrix} \\ [15pt]

    \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} &=& (-1)^{{j_1}+{j_2}+{j_3}} \begin{pmatrix}j_2 & j_1 & j_3 \\ m_2 & m_1 & m_3 \end{pmatrix} \\ [15pt]

    \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} &=& (-1)^{{j_1}+{j_2}+{j_3}} \begin{pmatrix}j_1 & j_2 & j_3 \\ -m_1 & -m_2 & -m_3 \end{pmatrix} \\ [15pt]

    \begin{pmatrix} j+1 & j & 1     \\ m & -m-1 & 1    \end{pmatrix} &=& (-1)^{{j-m+1}}{\left[\frac{(j-m)(j-m+1)}{(2j+3)(2j+2)(2j+1)}\right]}^{\frac{1}{2}} \\ [15pt]

    \begin{pmatrix} j+1 & j & 1     \\ m & -m & 1      \end{pmatrix} &=& (-1)^{{j-m-1}}{\left[\frac{(j-m+1)(j-m+1)}{(2j+3)(2j+2)(2j+1)}\right]}^{\frac{1}{2}} \\ [15pt]

    \begin{pmatrix} j   & j & 1     \\ m & -m-1 & 1    \end{pmatrix} &=& (-1)^{{j-m}}{\left[\frac{(j-m)(j-m+1)}{j(2j+1)(2j+2)}\right]}^{\frac{1}{2}}  \\ [15pt]

    \begin{pmatrix} j   & j & 1     \\ m & -m & 0      \end{pmatrix} &=& (-1)^{{j-m}}{\left[\frac{m^2}{j(j+1)(2j+1)}\right]}^{\frac{1}{2}}\\ [15pt]

\end{tabular}

\end{table} 
\end{document}

Which yields: enter image description here

1
  • Indeed, your idea is quite neat and elegant! Thanks a lot!
    – Strelok
    Nov 6 '17 at 10:46

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