The proof of the Wigner-Eckart theorem for irreducible tensor operators

I am reading through Wu-Ki Tung's Group Theory in Physics and I met a problem when going through the part of the Wigner-Eckart theorem for irreducible tensor operators. In the 4.3 part of the book, the author defines irreducible tensor operators as operators having the following transforming property($$U$$ is a general representation of the group $$G$$, while $$D^{\mu}$$ is the $$\mu$$'s irreducible matrix representation) : $$U(g) O_{i}^{\mu} U(g)^{-1}=O_{j}^{\mu} D^{\mu}(g)_{i}^{j}$$ and then he shows that the set of vectors $$U(g) O_{i}^{\mu}\left|e_{j}^{v}\right\rangle$$ ($$|e_{j}^{v}\rangle$$ is a irreducible set of $$\nu$$'s irreducible representation, corresponding to some physical states) transform according to the direct product representation $$D^{\mu \times \nu}$$:\begin{aligned} U(g) O_{i}^{\mu}\left|e_{j}^{v}\right\rangle &=U(g) O_{i}^{\mu} U(g)^{-1} U(g)\left|e_{j}^{v}\right\rangle \\ &=O_{k}^{\mu}\left|e_{l}^{v}\right\rangle D^{\mu}(g)^{k}{ }_{i} D^{v}(g)^{l}{ }_{j} \end{aligned}so we can express this set of vectors using the irreducible vectors of $$U^{\mu}\times U^{\nu}:\left|w_{\alpha l}^{\lambda}\right\rangle$$as follow (the coefficients of RHS are the Clebsch-Gordan coefficients,$$\alpha$$ distinguish different sets of the same irreducible representation) :$$O_{i}^{\mu}\left|e_{j}^{v}\right\rangle=\sum_{\alpha, \lambda, l}\left|w_{\alpha l}^{\lambda}\right\rangle\langle\alpha, \lambda, l(\mu, v) i, j\rangle$$ then, using the orthogonality of irreducible sets of different representations, the author states that we can proof the following formula for the matrix element $$\left\langle e_{\lambda}^{l}\left|O_{i}^{\mu}\right| e_{j}^{v}\right\rangle$$(the formula (4.3-4) of the book):$$\left\langle e_{\lambda}^{l}\left|O_{i}^{\mu}\right| e_{j}^{v}\right\rangle=\sum_{\alpha}\langle\alpha, \lambda, l(\mu, v) i, j\rangle\left\langle\lambda\left|O^{\mu}\right| v\right\rangle_{\alpha}$$where $$\left\langle\lambda\left|O^{\mu}\right| v\right\rangle_{\alpha} \equiv n_{\lambda}^{-1} \sum_{k}\left\langle e_{\lambda}^{k} \mid w_{\alpha k}^{\lambda}\right\rangle$$ is called the reduced matrix element, where $$n_{\lambda}$$ is the dimension of the $$\lambda$$'s irreducible representation.

then the author states that the through this formula the $$i$$- $$j$$- and $$l$$- dependence of the matrix element are all in the Clebsch-Gordan coefficients which can be determined through the representation theory.

My first question is, how can we derive the last formula through the process described above? I have tried but I can't get the right result:$$\left\langle e_{\lambda}^{l}\left|O_{i}^{\mu}\right| e_{j}^{v}\right\rangle=\sum_{\alpha, \lambda^{\prime}, l^{\prime}}\langle e_{\lambda}^{l}\left|w_{\alpha l^{\prime}}^{\lambda^{\prime}}\right\rangle\langle\alpha, \lambda^{\prime}, l^{\prime}(\mu, v) i, j\rangle \\=\sum_{\alpha , l^{\prime}}\langle e_{\lambda}^{l}\left|w_{\alpha l^{\prime}}^{\lambda}\right\rangle\langle\alpha, \lambda, l^{\prime}(\mu, v) i, j\rangle$$the second equality is due to the orthogonality of different irreducible sets.

My second question is, where does the $$\mu$$,$$\nu$$dependence of the reduced matrix $$\left\langle\lambda\left|O^{\mu}\right| v\right\rangle_{\alpha} \equiv n_{\lambda}^{-1} \sum_{k}\left\langle e_{\lambda}^{k} \mid w_{\alpha k}^{\lambda}\right\rangle$$come from? I think as follow: if we want to look up the Clebsch-Gordan coefficients from some pulished books, we must construct the set $$|w_{\alpha l}^{\lambda}\rangle$$ from $$O_{i}^{\mu}\left|e_{j}^{v}\right\rangle$$ in a specific way.

• Since I do not have this book, I cannot be sure why there is $\alpha$ in the CG coefficient. In wikipedia (en.wikipedia.org/wiki/Clebsch%E2%80%93Gordan_coefficients), CG coefficient is defined with $\langle j_1m_1j_2m_2|jm \rangle$. With your notation, it should be $\langle \nu j \mu i|\lambda l\rangle$. The wikipedia page of Wigner Eckart theorem also has this form of CG coefficients (en.wikipedia.org/wiki/Wigner%E2%80%93Eckart_theorem). Commented Aug 4, 2022 at 15:19
• $\alpha$ is used to distinguish different sets of the same irreducible representation, in the case of addition of angular momentum, each irreducible set only occurs once, so there isn't a label $\alpha$. Commented Aug 5, 2022 at 2:57

Thank you for your clarification on $$\alpha$$, so I can provide the answer more accurately.
The trick to deriving the last equation is to consider the product $$\langle e^{l}_{\lambda}|O^{\mu}_i|e^{\nu}_{j}\rangle$$ with \begin{align} \langle e^{l}_{\lambda}|O^{\mu}_i|e^{\nu}_{j}\rangle & = \langle e^{l}_{\lambda}|U(g^{-1})U(g)O^{\mu}_i|e^{\nu}_{j}\rangle \\ & = \langle e^{\color{red}{k}}_{\lambda}|(D^{\lambda}(g)^{\color{red}{k}}_{l})^*\sum_{\alpha,\lambda',l'}{\bigg(D^{\lambda'}(g)^{\color{red}{q}}_{l'}|\omega^{\lambda'}_{\alpha,{\color{red}{q}}}\rangle \langle \alpha,\lambda',l'(\mu,\nu)i,j \rangle \bigg)} \\ & = \sum_{\alpha,\lambda',l'}{\langle \alpha,\lambda',l'(\mu,\nu)i,j \rangle \bigg((D^{\lambda}(g)^{\color{red}{k}}_l)^*D^{\lambda'}(g)^{\color{red}{q}}_{l'}\bigg)\langle e^{\color{red}{k}}_{\lambda}|\omega^{\lambda'}_{\alpha,{\color{red}{q}}}\rangle} \end{align} where red labels are summed over as Einstein's convention. Now we integrate both sides over $$g$$, and use the orthogonality relation to come up with \begin{align} \langle e^{l}_{\lambda}|O^{\mu}_i|e^{\nu}_{j}\rangle & = \sum_{\alpha,\lambda',l'}{\langle \alpha,\lambda',l'(\mu,\nu)i,j \rangle{1 \over 2\lambda+1}\delta_{\lambda\lambda'}\delta_{{\color{red}{k}}{\color{red}{q}}}\delta_{ll'}\langle e^{\color{red}{k}}_{\lambda}|\omega^{\lambda'}_{\alpha,{\color{red}{q}}}\rangle} \\ & = \sum_{\alpha}{\langle \alpha,\lambda,l(\mu,\nu)i,j \rangle{1 \over 2\lambda+1}\sum_{k}{\langle e^{k}_{\lambda}|\omega^{\lambda}_{\alpha,k}\rangle}} \end{align}
This answers your first question, and about the second one, your thought is correct, the set of vectors $$\big\{|\omega^{\lambda}_{\alpha,k} \rangle\big|\lambda=|\mu-\nu|,\cdots,\mu+\nu, \ \text{and} \ |k| \leq \lambda\big\}$$ are constructed from $$\big\{O^{\mu}_i|e^{\nu}_{j}\rangle\big||i| \leq \mu, \ \text{and} \ |j| \leq \nu\big\}$$. Therefore, the dependence of $$\mu$$ and $$\nu$$ are not absent from the reduced matrix.