# Spherical Operators

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

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 $$$$. What does it mean? How can I compute the matrix elements?

• 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? Dec 15, 2020 at 11:37
• 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. Dec 15, 2020 at 16:42
• just to clarify since I can't edit: $J_{i} = J^{+},J^{-},J_{0}$ Dec 15, 2020 at 16:49