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I'm solving a question that required me to work with a Cartesian tensor, $A_i$, with Wigner-Eckart theorem.

The book was Sakurai Modern Quantum Mechanics with information given in related posts:

Ra. Decomposition of a Cartesian tensor

Rb. Irreducible form of Spherical tensor operators

Rc. Representing a reducible Cartesian tensor as a spherical tensor

The issue I have had was that the form of Wigner-Eckart theorem was given in spherical tensor. But $A_i$ was a cartesian tensor.

Qa. How to decompose $A_i$ into $T_{q}^{(k)} = Y_{l=k}^{m=q}(A_i)$? Is it a summation over index $m/q,l/k$? Does it have coefficients? And how to determine the rank of the tensor?

There was a projection theorem when $j=j'$, Eq 3.11.40 $$\langle \alpha', jm'|V_q |\alpha, jm\rangle = \frac{\langle \alpha', jm'|J\cdot V|\alpha, jm\rangle} {\hbar^2 j(j+1)} \langle jm' |J_q|jm\rangle$$

However, it was very confusing. From the reading, the book seemed to indicate that the subscript $V_q$ here indicate the spherical $m=q$.

Qb. Could projection theorem be used to deal with Cartesian tensor?

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    $\begingroup$ It depends upon what is the rank of $A$, if rank is one (it is a Cartesian vector), then you can easily transform from Cartesian to Spherical Tensor Basis $A_{+1}=-(A_x+i A_y)/\sqrt(2), A_{-1}=A_x-i A_y)/\sqrt(2), A_{0}=A_{z}$. If your Cartesian tensor is rank $2$, then it is reducible, which means that in the spherical tensor basis it is represented as the following sum $\mathbf{A} = \sum\limits_{k=0}^{2} \mathbf{A^{(k)}}$. You can find it easily, for instance, in web.archive.org/web/20190713054517/http://info.phys.unm.edu/…. $\endgroup$
    – Sl0wp0k3
    Commented Sep 30, 2023 at 10:00
  • $\begingroup$ For higher ranks of Cartesian tensors, you have to do it yourself, it is somewhat similar to the addition of quantum orbital angular momenta: you need to decompose higher ranks in terms of lower ranks using the formula for tensor products of irreducible spherical tensors, and for lower ranks (vectors (rank 1) and matrices (rank 2)) you know the answer already. However, the decomposition will look more complicated than that for rank 2, because there might be several spherical tensors of a given rank required. I recommend to search it on math stackexchange, there are many threads on this topic. $\endgroup$
    – Sl0wp0k3
    Commented Sep 30, 2023 at 10:06

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