
#1
Feb1408, 02:22 PM

Sci Advisor
HW Helper
P: 4,739

Is wondering if anyone knows if the modulus square of the double matrix element that arises in WignerEckart theorem obeys the same "rule" as the ordinary does, if the operator is hermitian:
[tex] <ajmMbj'm'>^2 = <bj'm'Majm>^2 [/tex] if M is hermitian. Is then : [tex] <ajMbj'>^2 = <bj'Maj>^2 [/tex] ?  I think it does, the WignerEckart theorem states: [tex] \langle njmT^k_qn'j'm'\rangle =\langle njT_qn'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 :) 


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