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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:
[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] ?
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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 :)
[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 :)