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?

  • 1
    $\begingroup$ It depends on how the tensor couples to the spin or the orbital angular momentum. For example, we know that electric dipole operator only couples to orbital angular momentum. Then you can decompose your J-reduced matrix element in terms of L-reduced matrix element combined with Wigner 6j symbol. See pg 7 of this document (steck.us/alkalidata/rubidium87numbers.1.6.pdf). $\endgroup$ – wcc Mar 17 '19 at 21:35
  • 1
    $\begingroup$ Also consult Chapter 7 in (atomoptics-nas.uoregon.edu/~dsteck/teaching/quantum-optics/…) for meaning of Wigner 6j symbol. It has to do with unraveling of three angular momenta (in this case, L, S, and your spherical tensor). $\endgroup$ – wcc Mar 17 '19 at 21:38
  • $\begingroup$ Thanks for the references, those look great. I'll have a chance to look more thoroughly in a few days $\endgroup$ – dsm Mar 18 '19 at 1:19

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.

| cite | improve this answer | |
  • $\begingroup$ Excellent, thanks! $\endgroup$ – dsm Mar 21 '19 at 0:32
  • $\begingroup$ I've now heard several great things about the Moskalev et. al. book, will definitely be starting that read. $\endgroup$ – dsm Mar 21 '19 at 0:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.