Is wondering if anyone knows if the modulus square of the double matrix element that arises in Wigner-Eckart theorem obeys the same "rule" as the ordinary does, if the operator is hermitian: [tex] |<ajm|M|bj'm'>|^2 = |<bj'm'|M|ajm>|^2 [/tex] if M is hermitian. Is then : [tex] |<aj||M||bj'>|^2 = |<bj'||M||aj>|^2 [/tex] ? --- I think it does, the Wigner-Eckart theorem states: [tex] \langle njm|T^k_q|n'j'm'\rangle =\langle nj||T_q||n'j'\rangle C^{jm}_{kqj'm'} [/tex] where [tex] C^{jm}_{kqj}[/tex] is a Clebsh gordan So I think things will work out, are someone sure about how these things work, please tell me :)