- Tensor Operators
1. Lie Group Representation
$D(\alpha)$ represents an element $g$ of the Lie group $G$, where $\alpha = (\alpha_1, \alpha_2, \ldots, \alpha_N)$, with $g(0) = e$, which implies $D(0) = 1$. Therefore,
$$D(d\alpha) = 1 + i d\alpha_a X_a$$
where,
$$X_a = -i \frac{\partial D(\alpha)}{\partial \alpha_a}$$
Also, an element $X$ in the Lie algebra can be expressed as
$$X = \alpha_a X_a = \boldsymbol{\alpha} \cdot \boldsymbol{X}$$
2. Lie Algebra and Subalgebra
If $J_a$ are elements of the Lie algebra of $G$, and $X_a$ are elements of a subalgebra, then
$$[J_a, X_b] = i\ \epsilon_{abc} X_c$$
This satisfies
$$[J_a, X_q] = [J^{(k)}_a]_{q'q} X_{q'}$$where $q’$ ranges from 1 to $2k+1$ in the subalgebra.
3. Transformation Properties
$$-i \frac{\partial}{\partial \alpha_a} e^{i \alpha_a J_a} X_q e^{-i \alpha_a J_a} \bigg|_{\alpha=0}$$ $$ = [J_a, X_q] = [J^{(k)}_a]_{q'q} X_{q'}$$Therefore,
$$e^{i \alpha_a J_a} X_q e^{-i \alpha_a J_a}$$ $$ = D^{\dagger}(R) X_q D(R) = [D^{(k)}(R^{-1})]_{q'q} X_{q'}$$4. Definition of Tensor Operators
Substituting $T^{(k)}_q$ for $X_q$, we define tensor operators as operators $T^{(k)}_q$ that satisfy
$$[J_a, T^{(k)}_q] = [J^{(k)}_a]_{q'q} T^{(k)}_{q'}$$ or$$[J_a, T^{(k)}_q] = [J^{(k)}_a]^*_{qq'} T^{(k)}_{q'}$$ Equivalently,$$D^{\dagger}(R) X^{(k)}_q D(R) = [D^{(k)}(R^{-1})]_{q'q} T^{(k)}_{q'}$$ or$$D^{\dagger}(R) T^{(k)}_q D(R) = [D^{(k)*}(R)]_{qq'} T^{(k)}_{q'}$$