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Double bar matrix element

by malawi_glenn
Tags: double, element, matrix
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malawi_glenn
#1
Feb14-08, 02:22 PM
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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 :)
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