Prove that

\[\left\{\begin{array}{ccc}j_{1} & j_{2} & J \\j_{1}^{\prime} & j_{2}^{\prime} & J^{\prime} \\k & k^{\prime} & 0\end{array}\right\}=\frac{(-1)^{j_{2}+j_{1}^{\prime}+k+J}}{\sqrt{(2 J+1)(2 k+1)}}\left\{\begin{array}{ccc}j_{1} & j_{2} & J \\j_{2}^{\prime} & j_{1}^{\prime} & k\end{array}\right\} \delta_{k k^{\prime}} \delta_{J J^{\prime}}\]

Use the results of Problem 30.1 .

Data from Problem 30.1

Prove that for $6 J$ symbols

\[ \left\{\begin{array}{lll} a & b & c \\ d & e & 0 \end{array}\right\}=\frac{(-1)^{a+b+c}}{\sqrt{(2 a+1)(2 b+1)}} \delta_{a e} \delta_{b d} \]

Note that since the $6 \mathrm{~J}$ symbol is invariant under interchange of columns or the interchange of upper and lower arguments in any two columns, this formula can be used to evaluate $6 J$ coefficients with a zero in any position. Hint: A $6 J$ recoupling coefficient can be expressed in terms of sums over products of $3 \mathrm{~J}$ coefficients [48],

\[ \begin{aligned} &\left\{\begin{array}{lll} j_{1} & j_{2} & j_{3} \\ \ell_{1} & \ell_{2} & \ell_{3} \end{array}\right\}=\sum(-1)^{-l_{3}-j_{3}-j_{1}-j_{2}-\ell_{1}-\ell_{2}-m_{1}-m_{1}^{\prime}} \\ & \times\left(\begin{array}{ccc} j_{1} & j_{2} & j_{3} \\ m_{1} & m_{2} & -m_{3} \end{array}\right)\left(\begin{array}{ccc} \ell_{1} & \ell_{2} & j_{3} \\ m_{1}^{\prime} & m_{2}^{\prime} & m_{3} \end{array}\right)\left(\begin{array}{ccc} j_{2} & \ell_{1} & \ell_{3} \\ m_{2} & m_{1}^{\prime} & -m_{3}^{\prime} \end{array}\right)\left(\begin{array}{ccc} \ell_{2} & j_{1} & \ell_{3} \\ m_{2}^{\prime} & m_{1} & m_{3}^{\prime} \end{array}\right), \end{aligned} \]

where the sum is over all $m$ and $m^{\prime}$.