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I had a question about Irreducible Spherical Tensor Operators and their relationship with the Clebsch Gordon coefficients.

Consider, then, a dyadic product $T_{ik} = X_{i}Z_{k}$ of two even parity rank 1 spherical tensors $X_{i}$ and $Z_{k}$, where $i,k = -1, 0, 1$. The questions is, how can one express the following matrix elements of $T_{ik}$ with respect to the $| j m \rangle$ states,

$\langle1 - 1| T_{-1 -1}| 11 \rangle$ and $\langle 1 0| T_{0 -1}| 11 \rangle$, in terms of the following quantities:

$\alpha_{1} $ = $\langle1 1| T_{0 0}| 11 \rangle$, $\alpha_{2} $ = $\langle1 1| T_{1 -1}| 11 \rangle$, $\alpha_{3} $ = $\langle1 1| T_{-1 1}| 11 \rangle$.

Here is my approach to this problem, but I seem to be getting lost in the mathematics, and I am starting to wonder if this approach is even right or not.

Construct the tensor using:

$T^{(k)}_{q}$ = $\sum_{q_{1}} \sum_{q_{2}} \langle 11; q_{1} q_{2}| 11; kq \rangle X_{q_{1}}Z_{q_{2}}. $

Let's explicitly work out at least one of the equations.

$T^{(2)}_{0}$ = $\sum_{q_{1}} \sum_{q_{2}} \langle 11; q_{1} q_{2}| 11; 20 \rangle X_{q_{1}}Z_{q_{2}}. $

Since $q_{1}, q_{2} = -1, 0, 1$. We can carry out the first sum and achieve the following:

$T^{(2)}_{0}$ = $\sum_{q_{1}} \langle 11; q_{1} -1| 11; 20 \rangle X_{q_{1}}Z_{-1} $ + $\sum_{q_{1}} \langle 11; q_{1} -1| 11; 20 \rangle X_{q_{1}}Z_{0}$ $ + \sum_{q_{1}} \langle 11; q_{1} 1| 11; 20 \rangle X_{q_{1}}Z_{1}$.

Now we can carry out the second sum. In theory, there should be nine terms in total, but that is simplified as most of the Clebsh-Gordon Coefficients (CGC) are zero because of the following know fact: $q = q_{1} + q_{2}$; otherwise, CGC are zero.

$\langle 11; 1 -1| 11; 20\rangle X_{1}Z_{-1}$ + $\langle 11; 0 0| 11; 20\rangle X_{0}Z_{0}$ + $\langle 11; -1 1| 11; 20\rangle X_{-1}Z_{1}$.

These CG coefficients can be looked up in the table and this equation is simplified to the following:

$T^{(2)}_{0}$ = $\frac{1}{\sqrt{6}} T_{1-1}$ + $\frac{2}{\sqrt{3}} T_{00}$ + $ \frac{1}{\sqrt{6}} T_{-11}$.

This is only one of the many equations one can develop. My "algorithm" to solve this problem is to develop all such equations, manipulate them, and get the desired matrix elements in terms of the desired quantities. However, I cannot seem to get to that point. May be I am on the wrong track here? May be I also need to use recursion relationships of tensor operators? Any help would be appreciated, especially if someone who would like to take a good shot at this question.

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  • $\begingroup$ Here's one hint: the operator $T_{ik}$ is reducible: it splits as $(3) \otimes (3) = (5) \oplus (3) \oplus (1)$. $\endgroup$ – Hans Moleman Dec 13 '19 at 19:14
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It seems what you want is the "reverse" expansion \begin{align} X_iZ_j=\sum_{Lk} T^L_{k} \langle 1i; 1j\vert Lk\rangle\, . \end{align} Of course if $X_i$ and $Z_j$ act in different spaces you will have to unwrap your $\vert JM\rangle$ into its $\vert \ell_1m_1\rangle \vert \ell_2m_2\rangle$ bits. The matrix element of your composite tensor then factors into a product of reduced matrix elements of the type \begin{align} \langle \ell'_1\ell'_2;J'\Vert XZ \Vert \ell_1\ell_2;J\rangle \propto W \times \langle \ell'_1\Vert X\Vert \ell_1\rangle\ell'_2\Vert X\Vert \ell_2\rangle \end{align} where $W$ is a Racah W coefficient.

This type of gymnastics is common in the usual angular momentum textbooks such as

Rose, M.E., 1995. Elementary theory of angular momentum. Courier Corporation,

Brink, D.M. and Satchler, G.R., 1968. Angular momentum.

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  • $\begingroup$ Thank you so much for your answer and advice. $\endgroup$ – Sohair Abdullah Dec 13 '19 at 20:53

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