Prove that for $6 J$ symbols [left{begin{array}{lll}a & b & c d & e & 0end{array} ight}=frac{(-1)^{a+b+c}}{sqrt{(2 a+1)(2 b+1)}} delta_{a e} delta_{b d}] Note that

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. 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 $m^{\prime}$.