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I have some familiarity with these things from a course I took in Sakurai a few years ago.

Cartesian tensors form a 3d irrep of $ \mathrm{SO}(3) $. The angular momentum operators $ J_+,J_-,J_z $ form a 3d irrep of $ \mathrm{SO}(3) $. The spherical tensors $ T^1_q, q=1,0,-1 $ also form a 3d irrep of $ \mathrm{SO}(3) $. These three 3d irreps of $ \mathrm{SO}(3) $ are all isomorphic and I have seen some of them described as the same.

Dyadic cartesian tensors (2nd tensor power) are also a well known case. They form a 9d reducible representation of $ \mathrm{SO}(3) $, see Decomposition of a Cartesian tensor. The dyadic tensor representation decomposes into irreducibles as $ 5+3+1 $. The $ 5 $ corresponds to $ T^2_q, q=0,\pm 1, \pm 2 $ (traceless symmetric part) the $ 3 $ corresponds to $ T^1_q, q=0,\pm 1 $ (antisymmetric part) and the $ 1 $ corresponds to scalar part.

What about a tensor product of three cartesian tensors (3rd tensor power)? This should be a reducible 27d representation reducing as $ (3*5)+(3*3)+(3*1) $ which should split into irreducibles as $ (7+5+3) + (5+3+1) + 3 $. There are many different copies of the 3d irrep in this decomposition. All the different "3"s are in some sense $ T^1_q,q= \pm 1, 0 $, but somehow they are all different?

For example, for the 1st tensor power we would write $$ T_1^1=J_+ \\ T^1_0=J_z \\ T^1_{-1}=J_- $$ But for the second tensor power we would write the $ 3 $ part of the decomposition into irreducibles (remember this part corresponds to antisymmetric tensors) as $$ T_1^1=J_zJ_+-J_+J_z=[J_z,J_+]=J_+ \\ T^1_0=J_+J_- - J_-J_+=[J+,J_-]=2J_z \\ T^1_{-1}=J_zJ_--J_-J_z-=[J_z,J_-]= -J_- $$ so the $ T^1_{q}, q= \pm 1, 0 $ span the same space for the 1st and 2nd tensor power. What is going on here?

In the 3rd tensor power there are three different copies of the irrep $ 3 $. Will all of these three copies of the $ 3 $ irrep also be "redundant" in the sense of each corresponding to the same space, the span of $ J_+,J_z,J_- $?

If not, what do these three different copies of the irrep $ T^1_{q}, q= \pm 1, 0 $ inside of the 3rd tensor power actually look like?

This question Quantum mechanics angular momentum spherical tensor components is relevant but only explicitly does the $ 7 $ part of the third tensor power (corresponding to $ T^3_q, q= \pm 3, \pm 2, \pm 1, 0 $) not any of the three $ 3 $ irreps.

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  • $\begingroup$ Do as your link suggests: Add three spin 1 states, Clebsch, and for your three spin 1s in the decomposition, parse out the symmetric and the two mixed-symmetry ones. It is a mere basis change. $\endgroup$ Oct 15, 2023 at 20:22

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