Selection rules with the Wigner-Eckart Theorem

Question

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?

Answer 1

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

Varshalovich, Dmitriĭ Aleksandrovich, Anatolï Nikolaevitch Moskalev, and Valerii Kel'manovich Khersonskii. Quantum theory of angular momentum. 1988.

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

Rowe, David J., and John L. Wood. Fundamentals of nuclear models: foundational models. World Scientific Publishing Company, 2010.

Various authors use various symbols such as the $W$ or $U$ symbols of Wigner and Racah respectively.