# 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?

• I do not completely understand your question. The reduced matrix element is not defined since the full matrix element is $0$. Can you clarify the context in which you might need to use the RME without the WE theorem? BTW, your are missing a conventional $1/\sqrt{2j_f+1}$ factor but this does not affect your question or my comment. – ZeroTheHero Jun 4 '18 at 5:03
• The problem is, that I have a model, where the full matrix element for any $m, m_i, m_f$ can be calculated independently and be shown that it is non-zero. For example, a model of a nucleus with $j=4$ transitioning electromagnetically to state with $j=0$. The state with $j=0$ does not necessarily be even the ground state, just some state with a zero total angular momentum. – user74200 Jun 4 '18 at 5:10
• If the model or the computation is wrong, but the transition probability is defined by reduced matrix element, how can I show, what is the transition probability? In the end, it does not matter if the actual full matrix element in the model is zero or not. The problem is, that the transition probability is defined by the reduced matrix element, which is according to W-E theorem undefined. That is a problem. – user74200 Jun 4 '18 at 5:20

1. 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 $0$.
2. 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 $j_f$.