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.

enter image description here

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!

  • 1
    $\begingroup$ One way of seeing it is by considering permutation symmetry of the states. Notice that for that case, states with $J=2,0$ are symmetric under permutation while the states with $J=1$ are antisymmetric. Hence the absence of $| m_1 = 0,m_2=0 \rangle$ in the $|J=1,M=0\rangle$ state. $\endgroup$ – secavara Jan 30 '18 at 21:24

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.


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.


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.