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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:
|<ajm|M|bj'm'>|^2 = |<bj'm'|M|ajm>|^2 if M is hermitian.
Is then :
|<aj||M||bj'>|^2 = |<bj'||M||aj>|^2 ?
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I think it does, the Wigner-Eckart theorem states:
\langle njm|T^k_q|n'j'm'\rangle =\langle nj||T_q||n'j'\rangle C^{jm}_{kqj'm'}
where C^{jm}_{kqj} is a Clebsh gordan
So I think things will work out, are someone sure about how these things work, please tell me :)
|<ajm|M|bj'm'>|^2 = |<bj'm'|M|ajm>|^2 if M is hermitian.
Is then :
|<aj||M||bj'>|^2 = |<bj'||M||aj>|^2 ?
---
I think it does, the Wigner-Eckart theorem states:
\langle njm|T^k_q|n'j'm'\rangle =\langle nj||T_q||n'j'\rangle C^{jm}_{kqj'm'}
where C^{jm}_{kqj} is a Clebsh gordan
So I think things will work out, are someone sure about how these things work, please tell me :)