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There is a given operator: $$T_j^m=\sum_{m_1,m_2}<j_1,j_2; m_1,m_2|j,m>X_{j_1}^{m_1}Y_{j_2}^{m_2}$$

Where $X_{j_1}^{m_1}$, $Y_{j_2}^{m_2}$ are spherical operators. I've to prove that $T_j^m$ is a spherical operator as well.

I think I don't really understand the projection $<j_1,j_2; m_1,m_2|j,m>$. What does it mean? How can I compute the matrix elements?

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the projection term is the matrix element of the change of basis matrix between the tensor product basis and the coupled basis. you can read about them here: https://en.wikipedia.org/wiki/Clebsch%E2%80%93Gordan_coefficients

calculating them isn't needed for your problem, though. recall what it means for a tensor to be a spherical tensor. what transformation rules and commutation relations must they satisfy? you've probably heard that spherical tensors transform similarly to angular momentum eigenstates. what does that mean?

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  • $\begingroup$ I know that when operator $J$ acts on spherical operator $O_j^m$ the commutator is: $[J_i, O_j^m] = \sum_{m'} [J_i^j]_{m'm}O_j^{m'} $. So simillary here, we get $[J_i, T_j^m] = \sum_{m'} [J_i^j]_{m'm}\sum_{m_1,m_2}<j_1,j_2; m_1,m_2|j,m>X_{j_1}^{m_1}Y_{j_2}^{m_2} $. Maybe I can do the same for each operator $X_{j_1}^{m_1}$ , $Y_{j_1}^{m_1}$ and check if the expression is similar? $\endgroup$
    – Konstantin Khrizman
    Dec 15, 2020 at 11:37
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    $\begingroup$ remember that spherical tensors "behave" under a a transformation $J_{i}$ the same way as if you applied J_i on a momentum eigenstate! what happens when you do $J_0|j,m>$? if an operator $\mathcal{O}^m_j$ follows the same result (except with the commutator instead of multiplication), then it is a spherical tensor. $\endgroup$
    – Tomka
    Dec 15, 2020 at 16:42
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    $\begingroup$ just to clarify since I can't edit: $J_{i} = J^{+},J^{-},J_{0}$ $\endgroup$
    – Tomka
    Dec 15, 2020 at 16:49

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