So this explanation (my first post on stackexchange!) is based on <PERSON> "Lie algebras in particle physics", chapter 4. Since $Q_{ij}$ is symmetric, real and traceless, it has 5 independent degrees of freedom. So it's possible to express $Q_{ij}$ into a 'spherical' basis $Q^s_l$, where $s=2$ in this case and $l$ takes on values -2, -1, 0, 1, 2. (For completness, a spherical tensor operator that transforms under the spin s representation of SU(2) is a set of operators such that: $[J_a,Q^s_l]=Q^s_m[J^s_a]_{ml}$).
The crux of the Wigner-Eckart theorem is that something like $Q^s_l |j,m,\alpha>$, which would physically correspond to electric effects of an atom which has an angular momentum, actually mathematically behaves like two angular momentum kets tensored together (recall adding angular momentum of particles in quantum mechanics). In terms of notation, this your quadrupole tensor operator acting on some state with total angular momentum squared j(j+1), and z angular momentum m. Here $\alpha$ parametrizes whatever other physics is going on in the system, for example non-perturbative quantum chromodynamic effects in the nucleus, which are super difficult to calculate.
The statement of the Wigner-Eckart theorem is that $=\delta_{m',l+m} * $.
On the LHS, you have the probability amplitude of measuring $Q^s_l|j,m,\alpha>$ to have total angular momentum J(J+1), z angular momentum $m'$ and with 'new' physics $\beta$ (which can be quite complicated). 976
On the RHS, the Wigner-Eckart theorem says that if you know the probability amplitude $$, which can be calculated for ANY of your $Q^s_l$ (calculated means an experimentalist has made a measurement or some hard working graduate student did the calculation), then using only group theory you know $$ for all the other $l$. All you need to do is calculate (or look up in tables) $$. This term originates from the crux of the theorem that the tensor operator times the ket behaves like two kets, which can be described in the basis of individual angular momentums |s,j,l,m> or in the basis of combined angular momentums $|J,l+m>$.
All this arises because the $Q^s_l$ operators form an irreducible representation, namely from some highest weight vector $Q^2_2$ in your example, you can get all the other $Q^2_l$ just by applying the lowering operator. So they are all related. Thus, the term $$ which contains all the nasty physics is universal constant within a given irreducible representation. So we don't need to calculate it for every $l$, only one of them (the easiest one).
Hope this helps! 976