Computing reduced matrix element without Wigner-Eckart theorem Lets have a problem: suppose we need to calculate reduced matrix element of some transition of a particle from some higher-order spin(or rather total angular momentum state, it does not really matter, it does not really matter how the particle does it - the only change when talking about a particular model is in the actual form of the hamiltonian, which is not important here - the problem is independent of the particular hamiltonian tensor model, it is just important, that the hamiltonian is rank 2 spherical tensor) state ($j=\{4,6,8,...\}$) to a ground state ($|j=0,m=0\rangle$) using a tensor Hamiltonian $\hat{H}=\hat{H}^2_m$. According to Wigner-Eckart theorem, it holds, that
$$\langle j_f,m_f|\hat{H}^j_m|j_i,m_i\rangle=\langle j_i, j, m_i, m|j_f,m_f\rangle \langle j_f||\hat{H}^j||j_i\rangle$$
However, if we set $j=2$, then the clebsh-gordan coefficient for any $j_i>2$ (and any $m, m_i$) is zero. In this case, we can not use Wigner-Eckart theorem, because we would be dividing by zero, hence we get no information on reduced matrix element of this transition through it. 
How can one compute a reduced matrix element of any transition if the CG coefficient is zero?
 A: I’m not sure I full understand but 


*

*There is no way a $j=2$ tensor can connect any states with $\Delta j=4$.  This is automatically forbidden by the integration over the angular part of the wavefunctions, i.e. $\int d\Omega (Y_{j_f,m_f}(\Omega))^* Y_{2m}(\Omega) Y_{j_im_i}(\Omega)=0$ if $j_f=j_i+4$, as in your example.  This is easily shown because $Y_{2m}\times Y_{j_im_i}$ will decompose into a sum containing at most $\Delta j_f=2$, so that the integration with a function where $\Delta j=4$ is immediately $0$.

*The reduced matrix elment does not depend only on $j_i$ and $j_f$, but formally also depend on any other quantum number needed to label your initial and final states.  In other words, you should really have
$$
\langle \alpha_f j_f m_f\vert H^j_m\vert \alpha_i j_i m_i\rangle
=\langle \alpha_f j_f \Vert H^j\Vert \alpha_i j_i\rangle 
\langle j_i j m_i m\vert j_f m_f\rangle
$$
where $\alpha$ is any remaining labels needed to completely specify your state. Therefore, the reduced matrix element for two final states with the same $j_f,m_f$ will be different if these states still differ by the remaining alpha labels.  As a result, you cannot compute the reduced matrix element for any $j_f=0$ and any $j_i=4$ and export this result to all other similar pairs of $j_f=0, j_i=4$ states. 


As a simple example, consider the set of 3d harmonic oscillator states.  The $N=2$ states have $\ell=0$ and $\ell=2$, and the $N=0$ state only has $\ell=0$.  Using $\vert N\ell m\rangle$ to denote states, the reduced matrix element $\langle N=0,\ell=0\Vert H^2\Vert N=2,\ell=2\rangle$ is not identical to the $\langle 2,0\Vert H^2 \Vert 2,2\rangle$ matrix element, even if both involve final states with $\ell=0$.  This is because the RME depends on the labels $N_i$ and $N_f$, not only on $j_i$ and $j_f$.
