# Selection rule on angular momentum quantum number for vector operators

According to the Wigner-Eckart theorem, the matrix element of a vector operator in the basis of eigenstates of the square of the magnitude of the angular momentum ($$J^2$$) and the z-component of the angular momentum ($$J_z$$) is proportional to a Clebsch-Gordan coefficient (I am aware that this theorem applies to all spherical tensor operators, but seeing as I'm not very familiar with these other tensor operators, it's best to stick with vector operators, though I'll take an answer discussing more general tensor operators as well). From this, it's concluded that the selection rule on the angular momentum quantum number is

$$\Delta j=0, \pm1 \tag{1}$$

since the relevant Clebsch-Gordan coefficients vanish whenever (1) is violated.

On Wikipedia, the theorem is derived by showing that the matrix elements satisfy the same recursion relations as the Clebsch-Gordan coefficients. The problem is that I don't see how this leads to the selection rules (1). Why can't there also be some other sets of quantities (apart from something directly proportional to the Clebsch-Gordan coefficients) that satisfy the recursion relations, but that don't vanish whenever (1) is violated? The selection rules on $$J_z$$ can be found by directly applying the commutators between $$J_z$$ and the vector operator, without having to worry about the recursion relations. But I've not seen a similar thing being done to $$J^2$$. So how does one arrive at the selection rule (1)?

• Please use MathJax to typeset equations, see physics.stackexchange.com/help/notation for details. Dec 30, 2021 at 20:47
• Should be fixed now. Thanks for pointing out. Dec 30, 2021 at 20:54

The key point is that $$T^{1}_{m_1}\vert \ell_2,m_2\rangle \tag{1}$$ transforms in exactly the same way as \begin{align} \vert 1,m_1\rangle\vert \ell_2,m_2\rangle \, .\tag{2} \end{align} You can see this because \begin{align} L_zT^{1}_{m_1}\vert \ell_2m_2\rangle &= [L_z,T^{1}_{m_1}]\vert \ell 2 m_2\rangle + T^{1}_{m_1}L_z\vert \ell_2 m_2\rangle\, ,\\ &=(m_1+m_2)T^1_{m_1}\vert\ell_2 m_2\rangle \end{align} and likewise the actions \begin{align} L_\pm T^{1}_{m_1}\vert \ell_2 m_2\rangle &=[L_\pm,T^1_{m_1}]\vert \ell_2m_2\rangle + T^{1}_{m_1} L_\pm \vert \ell_2m_2\rangle\, ,\\ &\sim L_\pm \left[\vert 1m_1\rangle \vert \ell_2 m_2\rangle\right] =\left[L_\pm\vert 1m_1\rangle\right]\vert\ell_2 m_2\rangle +\vert 1m_1\rangle\left[L_\pm \vert \ell_2m_2\rangle\right] \end{align} where by $$\sim$$ I mean the same factors with square roots come out, and of course $$[L_\pm,T^{1}_{m_1}]\propto T^1_{m\pm 1}$$.

If you believe that (1) does indeed transform in the same way as (2) does, then it is clear that, if (2) yields states with $$j$$ values ranging from $$\ell_2+1$$ to $$\vert \ell_2-1\vert$$, then the combination in (1) will also yields states with values of $$j$$ in that range. That’s basically the selection rule $$\Delta j=0,\pm 1$$.

Obviously if you do this not with a vector operator but a general tensor $$T^{\ell_1}$$, you will find the range of $$j$$ is just the same as the range in $$\ell_1\otimes\ell_2$$.

• A similar thing was shortly mentioned in Griffiths, but I guess I'm not entirely convinced that just because they transform similarly under a rotation, they have to have the same allowed values of $j$. Had I known why the combination of two angular momenta yield only those specific values of $j$, then perhaps that would have helped, but I imagine deriving the possible values of $j$ would require a much more advanced treatment that is past my level of understanding (right?). Dec 30, 2021 at 22:31
• not really. To obtain the possible values of $j$ just count the values of $m=m_1+m_2$ (with repetitions) and remember that $m_j$ runs from $-j$ to $j$. There is only one $m=\ell_1+\ell_2$ so $j=\ell_1+\ell_2$ must appear once. For this $j$ eliminate $m_j=j,j-1,\ldots,-j$ from your list of $m_1+m_2$ and start again: $m_1+m_2=j-1$ will appear once, etc. until you exhaust your list of $m_1+m_2$. Dec 30, 2021 at 22:48
• Griffiths is not so good for angular momentum coupling. I recommend the appropriate chapter in the (older) text by Gordon Baym. Dec 30, 2021 at 22:49
• Ah, I see. That makes sense. I guess also that since the application of $L_\pm$ and $L_z$ is the same for $T_q^k\left| l_1,m_1 \right>$ as for $\left| k,q \right>\left| l_1,m_1 \right>$, then the application of $L^2$ will also be the same. Then you just need to form the appropiate linear combinations of the states $T_q^k\left| l_1,m_1 \right>$ or $\left| k,q \right>\left| l_1,m_1 \right>$ to find eigenstates of $L^2$, which then will have the same eigenvalues for the two cases (and hence of course the same values of $l$). Dec 31, 2021 at 10:35
• Exactly. You build the combos using the Clebsch’s. Dec 31, 2021 at 10:42