Clebsch–Gordan coefficients

Remarks

The coefficients are real, and they vanish unless $M = m_1 + m_2$. For fixed $j_1, j_2$ they form an orthogonal matrix. Example: $j_1 = j_2 = \tfrac{1}{2}$, with $\lvert \uparrow \rangle = \lvert \tfrac12, \tfrac12 \rangle$ and $\lvert \downarrow \rangle = \lvert \tfrac12, -\tfrac12 \rangle$. Then $\lvert 1, 1 \rangle = \lvert \uparrow\uparrow \rangle$, $\lvert 1, 0 \rangle = (\lvert \uparrow\downarrow \rangle + \lvert \downarrow\uparrow \rangle)/\sqrt{2}$, and $\lvert 0, 0 \rangle = (\lvert \uparrow\downarrow \rangle - \lvert \downarrow\uparrow \rangle)/\sqrt{2}$. Check for $\lvert 0, 0 \rangle$: $J_+ \lvert \downarrow \rangle = \lvert \uparrow \rangle$ and $J_+ \lvert \uparrow \rangle = 0$, so $J_+ \lvert 0, 0 \rangle = (\lvert \uparrow\uparrow \rangle - \lvert \uparrow\uparrow \rangle)/\sqrt{2} = 0$, as for a vector of spin $0$. So $\langle \tfrac12\, \tfrac12; \tfrac12\, {-\tfrac12} \mid 0\, 0 \rangle = 1/\sqrt{2}$. Non-example: $(\lvert \downarrow\uparrow \rangle - \lvert \uparrow\downarrow \rangle)/\sqrt{2}$ also spans the spin $0$ part, but its coefficient of $\lvert \uparrow\downarrow \rangle$ is negative, so it breaks the Condon–Shortley convention. The Wigner $3j$ symbols are the same numbers up to a sign and a factor $\sqrt{2J + 1}$.