Working in the $|\alpha, j,m_j\rangle$ basis (denoting all irrelevant quantum numbers by $\alpha$), the Wigner-Eckart theorem tells us that the elements of a rank $k$ spherical tensor $T_q^{(k)}$ can be found with the Clebsch-Gordon coefficients:
$$\langle \alpha',j',m_j'|T_q^{(k)}|\alpha,j,m_j\rangle=\langle j,k;m_j,q|j,k;j',m_j'\rangle\langle\alpha',j'||T^{(k)}||\alpha,j\rangle.$$
From the Clebsch-Gordon coefficients we immediately know that the selection rules are
$$\Delta m_j=m_j'-m_j=q \\ |\Delta j|=|j'-j|\leq k,$$
and subject to parity we can usually restrict $|\Delta j|$ to either the even or odd integers. No problem there. My confusion comes when we explicitly include orbital angular momentum $l$ and spin $s$ into the above, represent our state in the $|n,l,s,j,m_j\rangle$ basis, and are interested in the elements
$$\langle \alpha',l',s',j',m_j'|T_q^{(k)}|\alpha,l,s,j,m_j\rangle.$$
I'm pretty sure the $\Delta m_j=q$ selection is entirely unaffected, but I'm confused about the second. We know that $|l-s|\leq j\leq l+s$, and I am tempted to define another parameter, call it $X$, where $|l-j|\leq X \leq l+j$, and state the modified selection rule as
$$|\Delta X|\leq k,$$
but I feel as though I'm somehow being redundant. Perhaps everything about orbital is already included in the $|\Delta j|\leq k$, and my comments of $X$ do not make any sense. So, my overall question: what are the selection rules for the matrix elements of a spherical tensor when working in the $|n,l,s,j,m_j\rangle$ basis?
Best Answer
I will write $$ \vert\alpha,\ell,s;j,m_j\rangle=\sum_{m_\ell m_s} C^{jm_j}_{\ell m_\ell; s m_s} \vert \alpha \ell m_\ell\rangle \vert \alpha s m_s\rangle\, , \tag{1} $$ and will assume that the tensor $T^{(k)}_q$ acts only on the orbital part, i.e. on the kets $\vert \alpha \ell m_\ell\rangle $. An expansion similar to (1) can be done for the bra $\langle \alpha',\ell',s';j',m'_j\vert $ and the Wigner Eckart theorem will produce a sum of the type \begin{align} \langle \alpha';\ell',s';j'm'_j\vert T^{(k)}_q \vert\alpha,\ell,s;j,m_j\rangle &=\sum_{m_\ell m_s m_{\ell'}} C^{jm_j}_{\ell m_\ell; s m_s} C^{j'm'_j}_{\ell' m'_\ell; s m_s}\delta_{ss'}\delta_{m_sm'_s} \frac{\langle \alpha' \ell'\Vert T^{(k)}\Vert \alpha \ell\rangle}{\sqrt{2\ell'+1}} C^{\ell' m_{\ell'}}_{kq;\ell m_\ell}\, \\ &=\frac{\langle \alpha j'\Vert T^{(k)}\Vert \alpha j\rangle}{\sqrt{2j'+1}}C^{j'm'}_{kq;jm} \end{align} The triple sum of Clebsch's on the right is actually proportional to the product of $ C^{\ell' m_{\ell'}}_{kq;\ell m_\ell}$ and a $6j$ symbol. Alternatively, one can multiply both sides by $C^{j''m''}_{kq;jm}$ and sum over $j'',m''$ to obtain a quadruple product of Clebsch's, proportional to a single $6j$ symbol. These summations can be found in
Either way, once these operations are done (they will involve permuting indices in the CGs ) one finally obtains \begin{align} \langle \alpha ' j' \ell' s'\Vert T^{(k)}\Vert \alpha j \ell s\rangle &=\delta_{ss'} \sqrt{\frac{2j'+1}{2\ell'+1}}U(s \ell j' k; j \ell') \langle \alpha' \ell'\Vert T^{(k)}\Vert \alpha \ell\rangle \tag{2}\\ &= \delta_{ss'}(-1)^{s+\ell+j'+k} \sqrt{(2j '+1)(2j+1)} \left\{\begin{array}{ccc}s&\ell&j\\ \ell &j&k \end{array}\right\}\langle \alpha' \ell'\Vert T^{(k)}\Vert \alpha \ell\rangle\, , \end{align} where $\left\{\begin{array}{ccc}s&\ell&j\\ \ell &j&k \end{array}\right\}$ is a $6j$ symbol. There are several version of Eq.(2), based on symmetries of the $6j$ symbols. The version given here is from
Various authors use various symbols such as the $W$ or $U$ symbols of Wigner and Racah respectively.