3-j symbol

Remarks

The symbol is $0$ unless $m_1 + m_2 + m_3 = 0$, since $J_z$ is additive on tensor products. It is unchanged by an even permutation of its columns and is multiplied by $(-1)^{j_1 + j_2 + j_3}$ by an odd permutation. The orthogonality relation reads $\sum_{m_1, m_2} (2 j_3 + 1) \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} \begin{pmatrix} j_1 & j_2 & j_3' \\ m_1 & m_2 & m_3' \end{pmatrix} = \delta_{j_3 j_3'} \delta_{m_3 m_3'}$.

Example: for $j_1 = j_2 = \frac12$ and $J = 0$, the invariant vector is $\lvert 0 0 \rangle = \frac{1}{\sqrt2}\left(\lvert \tfrac12 \tfrac12 \rangle \otimes \lvert \tfrac12, -\tfrac12 \rangle - \lvert \tfrac12, -\tfrac12 \rangle \otimes \lvert \tfrac12 \tfrac12 \rangle\right)$. So $\begin{pmatrix} \frac12 & \frac12 & 0 \\ \frac12 & -\frac12 & 0 \end{pmatrix} = \frac{(-1)^{0}}{1} \cdot \frac{1}{\sqrt2} = \frac{1}{\sqrt2}$. This agrees with the general value $\begin{pmatrix} j & j & 0 \\ m & -m & 0 \end{pmatrix} = \frac{(-1)^{j-m}}{\sqrt{2j+1}}$.

Non-example: $\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \end{pmatrix} = 0$, because $m_1 + m_2 + m_3 = 2 \ne 0$.

Physicists write the coefficients $\langle j_1 m_1, j_2 m_2 \mid J M \rangle$ also as $C^{J M}_{j_1 m_1 j_2 m_2}$.