Cohen-Tannoudji Complement AX: Addition of Angular Momenta missing vector I am currently going through the Cohen-Tannoudji quantum mechanics volume 2 textbook. I have reached the addition of angular momenta and am working on a complement in the book. The book is explaining how to find the |1,0> vector in the |J,M> basis. In this situation we have two particles both with orbital angular momenta equal to 1.

The book says it is interesting that a certain vector is not present in the |1,0> and while I understand mathematically why it is absent, I am not understanding why physically this would occur. Could anyone try and provide some reason why this is? I am assuming it has something to do with the singlet and triplet states, but I am unsure!
Thanks for any help in advance!
 A: This is sometimes due to symmetry arguments.  In the case of the $\vert 1,0\rangle$ state, the contribution from the coupled state $\vert 10\rangle_1\vert  1 0\rangle_2$ is clearly symmetric under permutation of particles but the states in the $L=1$ irrep arising from the coupling $(\ell_1=1)\otimes (\ell_2=1)$ must be antisymmetric, so must contain only state antisymmetric w/r to permutation of particles.  This justifies the absence of $\vert 10\rangle_1\vert  1 0\rangle_2$.
There are accidental zeros of the CG coefficients.  As the name implies, there appears to be non reason for them other than some accident of the summation.
A: From a tensor perspective, the $l=1$ vectors are:
$$ e^+ = -\frac{1}{2}(x + iy) $$
$$ e^0 = z $$
$$ e^- = \frac{1}{2}(x - iy).$$
These are eigenvector of $z$-rotations (on the unit sphere, they are the $l=1$ spherical harmonics-up to normalization). It's pretty clear that the dyad:
$$ e^0e^0 = zz $$
Now, decomposing a tensor into its invariant subspaces, the scalar part is:
$$ T^{(0)} = \frac{1}{3}\delta_{ij}T_{ii} \rightarrow \frac{1}{3}\delta_{ij}\propto Y_0^0,$$
where the last one has been evaluated for $zz$.
The vector part is antisymmetric:
$$ T^{(1)} = \frac{1}{2}(T_{ij}-T_{ji}) \rightarrow 0.$$
Per ZeroTheHero, $zz$ is totally symmetric under index interchange, and hence has no antisymmetric part.
The so-called natural-form (trace-free, symmetric), or pure rank-2 part of the tensor is:
$$ T^{(2)} = \frac{1}{2}(T_{ij}+T_{ji}) - T^{(0)}\rightarrow (2zz-xx-yy)/3 \propto Y_2^0.$$
The point here is that when constructing rank-$N$ Cartesian tensors, their (rotationally) invariant subspaces have the exact same form as the quantum spin combinations of $N$ vector particles--so understanding one can help with the understanding the other.
