To derive the expectation value of the nuclear tensor force, we recall that the product of two irreducible tensor operators is $$ W^{r}_{m_r}=\sum_{m_pm_q}\langle pm_pqm_q|rm_r\rangle T^{p}_{m_p}U^{q}_{m_q}, $$ and using the orthogonality properties of the Clebsch-Gordan coefficients we can rewrite the above as $$ T^{p}_{m_p}U^{q}_{m_q}=\sum_{m_pm_q}\langle pm_pqm_q|rm_r\rangle W^{r}_{m_r}. $$ Assume now that the operators \( T \) and \( U \) act on different parts of say a wave function. The operator \( T \) could act on the spatial part only while the operator \( U \) acts only on the spin part. This means also that these operators commute. The reduced matrix element of this operator is thus, using the Wigner-Eckart theorem, $$ \langle (j_aj_b)J||W^{r}||(j_cj_d)J'\rangle\equiv\sum_{M,m_r,M'}(-1)^{J-M}\left(\begin{array}{ccc} J & r & J' \\ -M & m_r & M'\end{array}\right) $$ $$ \times\langle (j_aj_bJM|\left[ T^{p}_{m_p}U^{q}_{m_q} \right]^{r}_{m_r}|(j_cj_d)J'M'\rangle. $$