Statement of the problem:
Given that a second-rank tensor operator transform as $$T'_{jk} = R_{jm}R_{kn}T_{mn}$$ where R is the three-dimensional rotation matrix, I need to find the commutation relations of T with the angular momentum operators.
My approach to solve the problem: I wrote that $$D^\dagger T_{jk}D = R_{jm}R_{kn}T_{mn}$$ is the transformation obbeyed, where D is the representation of rotations, that is $$D(R) = exp{-i\epsilon J/h}$$ where J is the generator of the transformations, that is the angular momentum operator. Expanding this, it gives
$$(1 - i\epsilon J + ...)T_{jk}(1 + i\epsilon J + ...) = R_{jm}R_{kn}T_{mn}$$ which will give in the end $$T_{jk} + i\epsilon[ T_{jk} , J] = R_{jm}R_{kn}T_{mn}$$.
Now, choosing the j's and k's and some direction to rotate, for example the z direction, I am able to write the right-hand side and it will give one of the commutators, for example:
$$T_{xy} + i\epsilon[T_{xy} , J_z] = \epsilon T_{xx} + T_{xy} - \epsilon T_{yy}$$ and it gives the commutation relation $$[T_{xy} , J_z] = -i(T_{xx} - T_{yy})$$ I dropped the h in the denominator some lines above. Proceeding this way I can find all the commutations I need.
It seems reasonable? It seems to me, but this commutation relation is some weird and nothing like the ones that I am used to. I just want to know if I can go on this way or is there some problem that I did not notice.
I read about tensor operator but everywhere uses the spherical representation, and I want some more general that do not go in some particular representation.
