Using the orthogonality properties of the Clebsch-Gordan coefficients, $$ \sum_{m_am_b}\langle j_am_aj_bm_b|JM\rangle\langle j_am_aj_bm_b|J'M'\rangle=\delta_{JJ'}\delta_{MM'}, $$ and $$ \sum_{JM}\langle j_am_aj_bm_b|JM\rangle\langle j_am_a'j_bm_b'|JM\rangle=\delta_{m_am_a'}\delta_{m_bm_b'}, $$ we can also express the two-body matrix element in \( m \)-scheme in terms of that in \( J \)-scheme, that is, if we multiply with $$ \sum_{JMJ'M'}\langle j_am_a'j_bm_b'|JM\rangle\langle j_cm_c'j_dm_d'|J'M'\rangle $$ from left in $$ \langle (j_aj_b) JM | \hat{V} | (j_cj_d) JM \rangle = N_{ab}N_{cd}\sum_{m_am_bm_cm_d}\langle j_am_aj_bm_b|JM\rangle\langle j_cm_cj_dm_d|JM\rangle $$ $$ \times \langle (j_am_aj_bm_b)M| \hat{V} | (j_cm_cj_dm_d)M\rangle, $$ we obtain $$ \langle (j_am_aj_bm_b)M | \hat{V} | (j_cm_cj_dm_d)M\rangle=\frac{1}{N_{ab}N_{cd}}\sum_{JM}\langle j_am_aj_bm_b|JM\rangle\langle j_cm_cj_dm_d|JM\rangle $$ $$ \times \langle (j_aj_b) JM | \hat{V} | (j_cj_d) JM \rangle. $$